Question

HCN has a pK_a = 6.2 x 10^-10. If a 50.0 mL of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH   a) after 8.00 mL base, b) at the halfway point of the titration, c) at the equivalence point.

Answers are below, I need to know how to get to them step by step/conceptually

Confirm: a) 8.49 , b) 9.21 c) 10.96

Answer 1

millimoles of HCN = 50 x 0.1 = 5

pKa = -log Ka

pKa = -log (6.2 x 10^-10)

pKa = 9.21

a) after 8.00 mL base

millimoles of NaOH = 8 x 0.1 = 0.8

HCN + NaOH ---------------------> NaCN + H2O

5 0.8 0 0

4.2 0 0.8 0.8

pH = pKa + log [NaCN/ HCN]

pH = 9.21 + log (0.8 / 4.2)

pH = 8.49

b) at the halfway point of the titration

at the halfway point of the titration pKa always equal to pH

pH = pKa

pH = 9.21

c) at the equivalence point

at equivalence point only salt remains

[salt] = 50 x 0.1 / (50 + 50) = 0.05 M

salt is the from strong base and weak acid . so pH > 7

pH = 7 + 1/ 2[pKa + log C] --------------------------------> formula for strong base weak acid salt

pH = 7 + 1/2 [9.21 + log 0.05]

pH = 10.96