Achieve Week 2 #7

[Ziyi Meng 2K]

Hi,

The question is "HClO is a weak acid ( Ka=4.0×10−8 ) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.071 M in NaClO at 25 degree Celsius?"

I understand how to solve the problem, but I don't understand the concepts behind it. What does Kb represent in this problem and why do we have to start finding it? I am just having a hard time making a connection between the Kb and the salt NaClO.

[Kurosh Zamiri 1I]

Hi! So in this question it asks for the pH of a solution of 0.071M NaClO, and we are given the Ka value of HClO.

Since Na+ in NaClO is a spectator ion, we can write the dissociation equation as: NaClO --> Na+ + ClO-

Therefore, we can relate ClO- and HClO, since they are conjugates. We know that in regards to conjugates, Ka*Kb = Kw. Thus, we use Kw to find the Kb value.

Given the Kb value, and the initial concentration of ClO, and the given equation: ClO- + H2O --> HClO + OH-, we can set up our ice table and solve for the OH- concentration. After finding this, we take the -log of it to find pOH. Using pH + pOH = 14, we can solve for pH. Hope this helps!

[Cassidy Chiong 2J]

Hi!

So we know that HClO acts as a weak acid. When it is put in water, it can form ClO- ions through the equilibrium reaction HClO+H2O<-->ClO-+H3O+. In this reaction, we know that ClO- is the conjugate base of HClO.

So, when we are given the compound NaClO, we know that it will dissociate into Na+ and ClO- ions when placed in water. Na+ ions will not react with water since Na+ is the conjugate acid of the strong base NaOH; thus we can ignore Na+. However, since ClO- is the conjugate base of a weak acid, it will react when placed in water to give us the equilibrium equation ClO-+H2O<-->HClO+OH-. In this reaction, we see ClO- acting as a base since it makes OH- ions. That is why we need to find the Kb of the reaction. Kb is used when we are determining the equilibrium constant of a base.