Question

(a) What is the pH of a solution that contains 0.100 M HClO and 0.300 M KClO? (b) If 50.00 mL of 0.100 M HCl is added to 1.00 L of the above buffer, what is the new pH?

(b) If 50.00 mL of 0.100 M HCl is added to 1.00 L of the above buffer, what is the new pH?

Answer 1

pKa for HClO = 7.53

From Henderson equation

pH = pKa + log {[Conjugate base] / [Acid]}

pH = 7.53 + log {[KClO] / [HClO]} = 7.53 + log ( 0.300 M / 0.100 M ) = 7.53 + log 3 = 7.53 + 0.477 = 8.00

pH = 8.00

b) Concentration of HCl added = 0.100 M = 0.100 mol/L

Volume of HCl solution added = 50.00 ml = 50.00 L / 1000 = 0.05 L

Number of moles of HCl added = 0.100 mol/L * 0.05 L = 0.005 mol

Concentration of KClO = 0.300 M = 0.300 mol/L

Concentration of HClO = 0.100 M = 0.100 mol/L

Volume of buffer solution = 1.0 L

Number of moles of HClO = 0.100 mol/L * 1.0 L= 0.100 mol

Number of moles of KClO = 0.300 mol/L * 1.0 L = 0.300 mol

After addition of HCl,

Number of moles acid = (0.100 + 0.005) mol = 0.105 mol

Number of moles of conjugate base = ( 0.300 - 0.005) mol = 0.295 mol

From Henderson equation,

pH = 7.53 + log ( 0.295 / 0.105) = 7.53 + log 2.81 = 7.53 + 0.4487 = 7.98

New pH after addition of HCl = 7.98

(b) question a and b are same.