Question

Consider the titration of a 100mL of 0.100 M HCN by .100M NaOH. (Ka for HCN is 6.2x10^-10) Calculate the pH after 50.0 mL of .100M NaOH has been added

Answer 1

after addition of 50 mL of .1 M NaOH, 50 mL of 0.1 M HCN will be completely neutralized to give NaCN and rest 50 mL will remain as HCN

so the moles of NaCN = 0.050 (L) X 0.1(M) = 0.005 moles of NaCN and moles of HCN = 0.005 moles.

therefore conc. of NaCN = conc. of HCN = 0.005 (mol) /0.150 (L) = 0.033M

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

pKa = -log Ka = 9.21

therefore pH = 9.21 + log 0.033/0.033 = 9.21

but HCN is acidic so the answer is wrong as pH should be below 7.

let's consider the equilibrium reaction

HCN + H2O H3O⁺ + CN^-

Ka = [H3O+][CN-] / [HCN] = x² / (0.33 - x)

Since the Ka value is very small, we can use an approximation that x will be very well.

That means that 0.33 - x is approximately 0.33, so we can re-write the equation as follows.

This saves us the effort of having to solve a quadratic.

6.2 X 10^-10 = x²/0.33

therefore x = (0.33 X 6.2 X10^-10)^1/2 = 1.43 X 10^-5

pH = -log[H⁺] = -log1.43 X 10^-5 = 4.8