Finding Ka of HCN from KCN solution at pH 9.70

[Aly]

hi forum,
would nebody be able to help me solve this problem?

0.18mol of potassium cyanide (KCN) was dissolved in 1.00L of a solution in which the pH was held constant at 9.70 at a temperature of 298K.

the equilibrium concentration of CN- was 0.13M.

Calculate Ka for the weak acid HCN.

ne help would be greatly appreciated. :)

 

[maverick280857]

What kind of an electrolyte is KCN? What do you think will happen when KCN is put in a solution the pH of which is "held constant"? What are the ions in solution? What are the constraints?

How about showing us what your reasoning/working is so far, first?

 

[thepatientmental]

Hi there! I can definitely help you with this problem. To calculate Ka for HCN, we can use the following equation:

Ka = [H3O+][CN-] / [HCN]

First, we need to find the concentration of HCN in the solution. Since KCN is a strong electrolyte, it completely dissociates into K+ and CN-. This means that the concentration of CN- is also 0.18 mol/L. Since the pH is 9.70, we can use the equation pH = -log[H3O+] to find the concentration of H3O+:

9.70 = -log[H3O+]
[H3O+] = 10^(-9.70)
[H3O+] = 2.3 x 10^(-10) mol/L

Now, we can plug these values into the Ka equation:

Ka = (2.3 x 10^(-10) mol/L)(0.13 mol/L) / (0.18 mol/L)
Ka = 1.67 x 10^(-10)

Therefore, the Ka for HCN is 1.67 x 10^(-10). I hope this helps! Let me know if you have any other questions. :)