Properties of Acids 1

 Hello fam! Yes,  we are almost there! Our discussion on acids would soon be wrapped up. I hope you are learning something? After our discussion on acids, we would discuss bases, pH, buffers, salts, titration curves and everything else that relates to these. We have a long ride ahead of us so relax and enjoy the journey! You would certainly agree with me that chemistry is indeed fun. Do I have a witness in the house?

Today I want to start talking about the properties of acids. Before I get into that, let me touch on something. An acid  could be dilute or concentrated. A concentrated acid is an acid that contains more acid molecules than water molecules. A dilute acid is an acid that contains more water molecules than acid molecules. Meanwhile, we need to understand the concept of concentration in chemistry.

I have noticed over my ten years of teaching science and mathematics that a lot of folks use terms that they do not know the meaning. This makes deep comprehension of topics in the subject difficult. My goal in this page is to help you to understand chemistry in-depth. Away with mere memorization of facts! Are we together? Let’s continue the journey.

Concentration refers to the amount of substance in a solution. If you want to cook rice, its common to measure the rice in cups before you add water and begin to cook. Lets say that Mr. Ruchin added 3 cups of rice to  3 L of water and Mr. Njoku added 6 cups of rice to 3L of water in different pots. Which pot contains more rice? I guess you know the answer. So Mr. Njoku’s pot is more concentrated (contains more rice) than Mr. Ruchin’s pot.

In chemistry, there are many different units of concentration. For the purpose of our discussion, we would stick with the unit of moles per liter usually shown as M. A 2M solution can be said to be more concentrated (contains more acid molecules) than a 0.5 M solution.

I have successfully established the fact that concentration has to do with the amount of substance that is added to a solution. Now a solution is composed of a solute and a solvent. The equation looks thus;

Solute + Solvent = Solution

If I get some sugar and add it to water, the solid sugar particles would soon disappear and then we would have something that looks more like the water but still contains the sugar now dissolved in the water. 

The solute is the solid sugar that I added, the solvent is the water into which the sugar was added and the clear liquid that results after adding the sugar is called a sugar solution. We shall describe the properties of solutions in more detail in a later topic.

If I add 3g of sugar to 2 L of water. Then in a different vessel I add 10 g of sugar to 1L of water, which solution is more concentrated? It must be the solution that contains more sugar in less volume of water. In which case, it’s the later solution.

Let me state that a solute could also be a liquid or a gas. Anything that dissolves in  liquid, no matter the state of matter to which it belongs is a solute and what is formed is a solution. In this sense, a solution is still formed when camphor dissolves in kerosene. A solution is also formed when chlorine gas is dissolved in water. Why does camphor not dissolve water? This would be covered in a series that would come up later.

Is the concentration thing now clear to all? I believe it is. If you still have questions, reach out to me via comments section, WhatsApp, messenger, calls or text messages to our dedicated phone line (+234-8090526258).

Lets stop here for today. The discussion would continue tomorrow as we look at the physical and chemical properties of acids.

Solution to yesterday’s exercise;

Question; A weak acid with a dissociation constant 1.54 * 10^-5 is 1.26% dissociated. What is the concentration of the hydrogen ions in the solution?

Solution;

Yesterday we said that the percentage acid dissociation is given by;

α = √Ka/C * 100

In this case, we have that;

α = 1.26%

Ka = 1.54 * 10^-5

The question is asking us about the hydrogen ion concentration thus we must first find the initial concentration of the acid from the percent dissociation formula and then apply the equilibrium formula to obtain the concentration of the hydrogen ion H^+. We need to know that we are using the dissociation equation; 

HA(aq) <-->H^+(aq) + A^-(aq)

Ka = [H^+] [A^-]/[HA]

Back to getting the initial concentration of the acid;

1.26 = √1.54 * 10^-5/C * 100

1.26/100 = √1.54 * 10^-5/C

(1.26/100)^2 = 1.54 * 10^-5/C

C = 1/(1.26/100)^2 * 1.54 * 10^-5

C = 0.097 M

Given that;

[H^+] = [A^-] = x

And

[HA] =  0.097 M

Setting up the ICE table we have that;

            HA(aq)  <--> H^+(aq) + A^-(aq)

I        0.097               0                0

C       -x                    +x              +x

E    0.097 - x             x               x

We Recall that 

Ka = [H^+] [A^-]/[HA]

Ka = x^2/[HA]

1.54 * 10^-5 = x^2/0.097 – x

1.54 * 10^-5 (0.097 – x) = x^2

1.49 * 10^-6 - 1.54 * 10^-5x = x^2

X^2 +1.54 * 10^-5x - 1.49 * 10^-6 = 0

X = 0.0012

Again; 

[H^+] = [A^-] = x

[H^+] = 0.0012 M

The concentration of the hydrogen ion in the solution is 0.0012 M. Hope its clear? Reach out to me if there is any confusion at any step of the way.

Hope you enjoyed our lesson today?

Follow  our page  and share!

Do you have assignments, term papers, projects and difficult homework questions? Contact us for help now! We also offer home and remote tutoring services to students at all levels in science subjects

Our articles and other important details are available at our blog:

www.tolearnchem.blogspot.com