Chemistry Made Easy

 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

 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/[

 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

 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

 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