Calculate the pH of a solution

Question

Calculate the pH of a solution that results from mixing 43 mL of 0.13 M HCN(aq) with 48 mL of 0.11 M NaCN. The K_a value for HCN is 4.9 x 10^-10.

My answer was 4.71

The correct answer is 9.29

Why is 9.29 the correct answer. I thought you would subtract 14-9.29???

Answer 1

I don't know how you did the problem, so cannot tell you where you went wrong. Have a look at this approach.

Henderson Hasselbalch equation...

pH = pKa + log [salt]/[acid]

pKa = -log Ka = -log 4.9x10^-10 = 9.31

Now we need to find the final concentrations of the salt and the acid...

Final volume = 43 ml + 48 ml = 91 ml

Using M1V1 = M2V2, we can find final concentrations of both salt and acid.

[Salt]: (48 ml)(0.11 M) = (91 ml)(x M) and x = 0.058 M salt

[Acid] = (43 ml)0.13 M) = (91 ml)(x M) and x = 0.061 M acid

pH = 9.31 + log (0.058/0.061)

pH = 9.31 + (-0.022)

pH = 9.29