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pH range of color change for indicators

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 The pH range of color change for an indicator is the range of pH values over which the indicator changes color. Different indicators have different pH ranges of color change. Let me quickly remind you of something that we established in the last lesson. We saw that the pH at which the color change occurs must be the point at which pH = pKin (check the page timeline for details of the previous lesson). At this point, the color change is observed in the system and the color change usually spans through two pH units. It then follows that if we know the pKin of the indicator, we can be able to establish the pH range by counting two pH units around the value of pKin of the indicator. Let me also state that the pKin is the same as the pKa of the indicator. Let us look at phenolphthalein or instance; the pKa (pKin) is 9, the pH range is 8 – 10. For methyl orange; the pKa(pKin) is 3.5 while the pH range is 3 – 5. Finally let me say this; the pH range of color change for an indicator is determ
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How indicators work

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 Hello everyone, today we are going to look further into the way indicators work. We must say that indicators are weak organic acids or bases. Let us recall that a weak acid or base does not dissociate completely in solution. We can thus write a dynamic equilibrium that is set up when the indicator molecule is dissolved in water. The color changes of solutions that occur in acidic, basic or neutral media when an indicator is added to each stems from the fact that the color of the protonated and deprotonated forms of the indicator are different. It is safe to say that the degree of dissociation of the indicator depends on the concentration of the hydrogen ion in the medium (which is also the pH) of the solution. Let us consider the indicator HIn which can have a deprotonated form In-. Then we have that; HIn > H^+ + In- It follows that Kin is the equilibrium constant for the dissociation of the indicator as in; Kin = [H^+] [In-]/[HIn] We can rearrange the equation to have that; Kin/[
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Introduction to Indicators

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 Hello fam! Welcome back from the short break. I believe that we are all enjoying the summer holidays. We will now continue from a new topic after we discussed about salts last week. Today, we want to look at the concept of indicators. We need to know that an indicator is a substance that changes color in the presence of an acid or a base.  The idea of the color change is the first thing that you have to bear in mind about the indicators. Many of us have at one time or another titrated an acid against a base or vice versa.  The point at which the acid and base have completely reacted in accordance with the stoichiometry of the reaction equation is called the equivalence point. Let me give you a typical example; Look at the reaction; 2NaOH(aq) + H2SO4(aq) -- Na2SO4(aq) + 2H2O(l) This reaction would be complete when two moles of sodium hydroxide react with exactly one mole of sulfuric acid. In other words, the amount of titrant added is just enough to neutralize the analyte solution. Ho
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Solution to Self test Questions on water of Crystallization.

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 Hello fam! Due to some technical issues we could not post our lesson yesterday. However, today, we are going to consider the solutions to the self test problems that we gave on Tuesday. This is to help everyone to understand the concept better. Self test problems; 1) Determine the value of X in the formula of washing soda crystals given the equation below; Na2CO3.xH2O -> Na2CO3 + xH2O 28.6g                         10.6 g 2) A 46.1 g sample of hydrated potassium dichromate K2Cr2O7.xH2O was heated to give 35.3 g of anhydrous salt. What is the value of x? Solution; Question 1; From the question; Mass of hydrated salt = 28.6g   Molar mass of hydrated salt = 2(23) + 12 + 3(16) + x[2(1) + 16] 106 + 18x Mass of anhydrous salt =  10.6 g Molar mass of anhydrous salt = 2(23) + 12 + 3(16) = 106 g/mol Again; Number of moles of hydrated salt = Number of moles of anhydrous salt But number of moles = mass / molar mass Number of moles of hydrated salt = 28.6 g/106 + 18x g/mol Number of moles of hy
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Calculations involving water of crystallization

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 Hello everyone! Welcome to the month of August. Just like I announced yesterday, we are going to look at a couple of mathematical problems that involve water of crystallization of a salt. In all these problems, we can see that the hydrated salt is heated to produce the anhydrous salt. Example 1; Find x in the formula MgSO4.xH2O if 5.0 g of magnesium sulfate was dehydrated and 2.44 g remained. Solution; Molar mass of hydrated salt = 24 + 32 + 3(16) + x[2(1) + 16] = 104 + 18x g/mol Molar mass of anhydrous salt = 24 + 32 + 3(16) = 104 g/mol We know that; Number of moles of hydrated salt = Number of moles of anhydrous salt But number of moles = mass/molar mass Number of moles of hydrated salt = 5 g/104 + 18x g/mol Number of moles of anhydrous salt = 2.44 g/104 Hence; 5 /104 + 18x = 2.44/104 5 * 104 = 2.44(104 + 18x) 520 = 253.76 + 43.92x X = 520 – 253.76/43.92 = 6 The formula of the hydrated salt is MgSO4.6H2O Example 2; A 0.970 g sample of a hydrated copper II salt, CuX2.6H2O was heated
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Water of Crystallization

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 Hello fam! Today we would be looking at the water of crystallization of a salt. It is common to refer to some salts as hydrates. A hydrate is a salt that has its water of crystallization attached to its structure. The term water of crystallization refers to the number of water molecules that combine chemically, in definite molecular proportion with a salt to form a crystalline structure. The number of water molecules attached to the crystal lattice of the salt is usually written alongside the chemical formula of the salt by placing a dot in between the usual formula of the salt and the number of molecules of water attached to the salt lattice. Examples of salts having water of crystallization are; • Na2CO3.10H2O – Sodium carbonate decahydrate • CuSO4.5H2O  - Copper sulfate pentahydrate • CaSO4.2H2O – Calcium sulfate dihydrate  It is important to note that the presence or absence of the water of crystallization affects the properties of the salt. For instance, CuSO4.5H2O is blue in col
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Hydrolysis of Salts

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 Yesterday we discussed the uses of salts and today, I want us to discuss what happens when we dissolve a salt in water. In order to keep things as simple as possible, I want to say that when we dissolve a salt in water, the salt is interacting with the water in a unique way. We can say that the salt is reacting with the water.  When a substance is reacting with water, we say that the substance is undergoing hydrolysis. As such, the reaction of a salt with water is called hydrolysis of the salt. The nature of the solution that is formed when a salt is dissolved in water depends on the way the salt is hydrolyzed in solution. Generally speaking; - When a salt is formed from a strong acid and a strong base, the salt solution is neutral. - When a salt is formed from a strong acid and weak base, the salt solution is acidic - When a salt is formed from a weak acid and a strong base, the salt solution is basic. Let us now see the hydrolysis of a few salts to substantiate our argument;  - Cons
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Preparation of Salts

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 Today, our focus would be on the preparation of salts. We are concerned about how we can make salts. I want to remind you again that a salt is made when hydrogen ion in an acid is replaced by another cation. Let us now consider some of ways by which salts can be made; 1) Reaction of a dilute acid and metal; A salt can be formed when a single replacement reaction occurs between a dilute acid and a metal. The metal used here must be above hydrogen in the activity series. For example; Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g) This kind of reaction is most times given as one of the chemical properties of acids. 2) Neutralization reaction: As I have explained previously, a neutralization reaction occurs between an acid and a base to form a salt and water only. This is one of the most important ways by which soluble salts are formed. For example; KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) H2SO4(aq) + CuO(s) -> CuSO4(aq) + H2O(l) 3) Dilute acid and carbonate: One of the chemical properties
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Uses of Salts

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 Hello, I am glad to be with you again today as we continue to look at the topic, “Salts” in chemistry. Yesterday, we recapped the highlights of our study on complex salts and their different uses. Today we shall look into the broader picture and discuss the uses of salts generally. The different uses of salts are as follows; 1) Salts are used to lower the melting point of ice: It is common during winter to see ice all over the streets making vehicular traffic on the roads quite difficult. In such cases, it is not uncommon to see ice sprinkled on the road in order to lower the melting point of ice and clear the snow off the road. The property applied here is the ability of impurities (in this case salt particles) to lower the melting point of a pure substance (in this case ice). NaCl and CaCl2 are commonly used for this purpose. 2) Salts are used in the process of salting out: I have earlier discussed the process of saponification. In this process, sodium hydroxide is added to fat/oil
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Summary of Lesson on Complex Salts

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 In the past two days, we have been looking at the concept of complex salts and I want us to tie it up today so that we can look at other topics also. I will recap the highlights of our discussion to make it easier for you to remember the key points that we have mentioned in the last two days. I told you that a complex salt contains a complex cation or complex anion or both. The complex cation or anion does have a coordination sphere in which ligands which may be anionic or neutral are bonded to a central metal atom/ion covalently. This is called the coordination sphere of the complex ion.  The number of ligands found in the coordination sphere is the coordination number of the central atom/ion. The complex ion often have ions that balance the charges and these are called counter ions and are located outside the coordination sphere of the complex ion. The bonding between the complex ion and the counter ions is entirely ionic in nature. The geometry of the complex ions depend on the num
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Bonding in Complex Salts

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 Hello! yesterday, we started looking at the complex salt which we called tetramine copper II chloride with the formula; [Cu(NH3)4]Cl2. We did say that this is a complex salt because of the presence to the complex cation [Cu(NH3)4]^2+. I want to state that if we had a complex anion, the salt would also be a complex salt. A typical example of this is; K4 [Fe(CN)6]. The complex ion is  [Fe(CN)6]4- which is anionic and the counter ions in this case are potassium cations. I told you that we will look at the bonding in the complex salt [Cu(NH3)4]Cl2 and why it does not produce simple ions in water. I will try to explain the basics of the concept today as we would still look at coordination chemistry in more detail sometime later.  Now, the complex salt [Cu(NH3)4]Cl2 has both ionic and covalent bonds. The four ammonia molecules are attached to the central copper II ion by a coordinate covalent bond(see the page timeline for an explanation of what a coordinate covalent bond is). We showed the
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Introduction to complex salts

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 Hello fam! Welcome to today’s lesson. We would focus our study today on trying to understand what we mean by a complex salt. I advise that you look at the lessons on June 18 and July 6, 2023 found on the page timeline for a detailed explanation of the idea of the coordinate bond. I will try to give a little introduction here but looking at the lessons I quoted above are indispensable to making the most out of this lesson.  A coordinate bond is formed when ONLY ONE of the bonding species supply the electron pair that is shared in the covalent bond. In other words, a coordinate bond must involve a  Lewis acid (electron pair acceptor) and a Lewis base, (electron pair donor). Let us look at a typical example. Consider the formation of [Cu(NH3)4]Cl2. Remember that the ground state electron configuration of the copper II ion is written as; [Ar]3d9 4s0.  This implies that we have empty 4s and 4p orbitals that can participate in dsp2 hybridization (forming a square planar complex) along with
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Double Salts

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 Hello fam! Welcome to today’s lesson. We would focus our study today on trying to understand what a double salt is. While growing up in Nigeria, we had this substance that was used to wash snail before it is cooked. The substance was called “alum”. The substance was used to remove the snail slime so that the snail could easily be prepared in the pot. This same alum was used sometimes for water purification. I remember a number of times that my mother would fetch water from untrustworthy sources due to scarcity of portable water in some parts of Nigeria and she would add this substance to coagulate the particles in the water and cause them to flocculate and subsequently be filtered out.  My mother was a practicing chemist without even knowing it! It was years latter when I gained admission into the University of Nigeria Nsukka and started studying about water purification as a student whose major was Chemistry that I understood what my mother was doing back then when I was a child. Tha
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Types of salts 1

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 Hello fam! Welcome to today’s lesson. I am glad to be your host again as we learn chemistry together. Yesterday, we explained the meaning of the term “salt” and I believe that everyone got what we were tying to explain. Today we would start looking at the types of salts. We would continue with this tomorrow but today we are going to look at three kinds of salts for a start.  On a lighter note, I believe our lesson yesterday cleared up the erroneous idea that the only kind of salt that exists is the kind of salt that is used to cook in the kitchen. That one is just sodium chloride enriched with other ingredients to improve our health. We commonly refer to it as table salt. So lets go down to business. What are the types of salts? 1) Normal salts: These are salts that do not contain hydrogen or hydroxide ions in the salt formula. They are usually formed when a neutralization reaction occurs between a strong acid and a strong base such that the reaction is COMPLETE. An incomplete reactio
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Introduction to the Idea of Salts

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 Yesterday, we concluded a rich discussion on the pH scale and we solved nine mathematical problems that relate to pH, pOH, pKb, pKa etc then I gave you one problem to figure out on your own. I am still waiting for your response to that self test question.  Just as I announced yesterday, we are beginning a discussion of the concept of salts today. I twice mentioned when we were looking at the properties of acids and bases that formation of salts is a chemical property of both acids and bases. In fact, a salt is formed BY THE REPLACEMENT OF THE HYDROGEN IONS IN AN ACID BY METALS IONS OR OTHER POSITIVELY CHARGED SPECIES SUCH AS AMMONIUM IONS. The preceding phrase was written in upper case for emphasis. My concern in today’s lesson is to explain this phrase so that we can all be on the same page as we proceed in this discussion. I would list five salts and their corresponding acids and see if you can agree with the preceding paragraph after looking at the list; Salt name               Sal
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Calculations on pKa and pH

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 Hello! I heartily welcome you all to today’s lesson. We are going to look at the last two calculations that pertain to pH, acids and bases today and tomorrow we would commence our discussion on salts to pave way for a discussion on buffers. I hope all the topics so far are making sense? Your feedback would be highly appreciated. Example 8; What is the pH of 0.2 M solution of sodium acetate. The pKa of acetic acid is 1.8 * 10^-5. Solution; We have that the hydrolysis of sodium acetate occurs as follows; CH3CH2COONa(aq) <-> CH3CH2COO^-(aq) + Na^+(aq)    CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq) We can see that that the acetate ion acted as a base hence it accepted a proton. We would have to obtain the Kb Kb = 1 * 10^-14/Ka Kb = 1 * 10^-14/1.8 * 10^-5 Kb = 5.6 * 10^-10 Setting up the ICE TABLE, we have that;          CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq) I            0.2                                                      0                       
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Further problems on pH

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 Hello dear! Welcome to today’s lesson. Do not hesitate to reach out to us via Whatsapp, Messenger or comments on the page if you need clarification on any of the problems that have been solved here. Today, we would continue as usual with mathematical examples on pH and we would be done with all these tomorrow. On Saturday, we shall begin to discuss about salts which would ultimately culminate in a discussion about  indicators as well as the concept of buffers. Stay tuned as we take this long ride! Note; If the ICE tables are not displaying correctly on your device, consider flipping the device to landscape or using another device having a larger screen. Example 6; The pKb of pyridine is 8.77. What is the pH of a 0.010 M aqueous solution of  pyidine. Solution; We  must recall that pyridine is a weak base so the use of an ICE table is indispensable when we try obtain the equilibrium concentrations that would go into the equilibrium equation. We have to set up the ICE table as follows;  
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More Mathematical Problems on pH

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 Hello fam! Welcome to today’s class, we are continuing along the path that we started a few days ago. We have been looking at different problems that have to do with the pH and today we shall look at some more problems. Example 3; A solution has a pOH of 6.50 at 50oC. What is the pH of the solution? Solution; We know that; pH + pOH = 14 pH = 14 – 6.50 pH = 7.5 Example 4; Calculate the pH and pOH of 0.02 M H2SO4. Solution; We know that we would first find pH because it is an acid; pH = -log(0.02 M) pH = 1.69 pOH = 14 – 1.69 =12.31 Example 5; Calculate the pH and pOH of a solution of 3 * 10^-2 M Solution; We know that we would first find pOH because it is an base; pOH = -log(3 * 10^-2) = 1.5 pH = 14 – 1.5 = 12.5
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Calculations in pH and pOH I

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 Hello fam! You must have noticed that there was no lesson updated on our page yesterday. This is due to the fact that I have been ill since the weekend and I am currently taking as much rest as possible. I am recovering gradually. I was advised to take out the whole of yesterday to rest hence I was of duty and couldn’t deliver the lesson for yesterday.  Like I said in on Sunday, we shall look at mathematical examples of pH, pOH, pKa and pKb. Let’s get started with two examples today. Example 1; Find the pH of 0.0025 M HCl solution Solution; The first thing that must come to our mind is that HCl is a strong acid hence it completely dissociates in water. Since it dissociates completely, we do not need an ICE table in order to determine the pH. The ICE table would otherwise be necessary when we are dealing with a weak acid. We now use the formula; pH = -log[H^+] Where we have; [HCl] = [H^+] = [Cl^-] =  0.0025 M  pH = -log(0.0025 M) pH = 2.6 Example 2; What is the hydrogen ion concentrati
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Further introduction to the concept of pH

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 Hello, yesterday, we started off looking at the meaning of the term pH in chemistry and we agreed that the pH is defined as the negative logarithm of the hydrogen ion concentration. In modern chemistry, the p stands for "the negative decimal(base ten) logarithm of". Previously we saw that pKa means the negative logarithm of the acid dissociation constant. Also, pKb means the negative logarithm of the base dissociation constant. Do you now see that our definition of the term “p” in chemistry holds true always? Closely related to the pH is another term called the pOH. The pOH is the negative logarithm of the hydroxide ion concentration. Thus, while pH and pKa applies to acids, pOH and pKb applies to bases. Is that making sense? Without much ado, we have to bear in mind that the pH tells us how acidic a substance is. It is the indicator of the degree of acidity or alkalinity (basicity) of a substance. Generally, the pH of substances can be arranged in a scale of 1 – 14 and this
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Meaning of pH

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 Hello, In today’s lesson, we would discuss about a term that has come up once before in our discussion and I promised to delve into more details about it at a latter time.  Hence, we are going to look at the meaning of the term pH. In our lesson today, we would try to understand the meaning of pH. The term "pH" was first described by Danish biochemist Søren Peter Lauritz Sørensen in 1909. pH is an abbreviation for "power of hydrogen" where "p" is short for the German word for power, potenz and H is the element symbol for hydrogen. The H is capitalized because it is standard to capitalize element symbols. The abbreviation also works in French, with “pouvoir hydrogen” translating as "the power of hydrogen.” In today’s language, we understand the pH to be the negative logarithm of hydrogen ion concentration. That is; pH = -log(H^+) Or pH = -log(H3O^+) In tomorrow’s lesson, we shall try to understand the pH scale and how the scale works. See you then! Ho
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Further uses of bases

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 Hello! Welcome to today’s lesson. It’s indeed very awesome that we have made this much progress learning together as an online community. We look forward to doing greater things together. As we explained yesterday, today we would look at more uses of bases and you would certainly find this very interesting. Yesterday, we saw the important role that bases play in the body when the acidity in the stomach goes above the normal level. It would interest you to know that in petroleum refining, an important step is to wash the raw crude with strong acids such as oleum. This serves the purpose of removing impurities in the crude. If we charge this highly acidified crude into the refining tower, the acid in the crude could corrode the metal used to make the tower and badly damage it. Hence an equally strong base such as NaOH is used to wash the crude oil so that the acid is neutralized before charging the crude oil into the refining tower. All of you use soap right? Let me tell you how soap is
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