Question

Calculate the pH during a titration when 7.50 ml of 0.15 M HCl solution has reacted with 19.29 ml of 0.15 M NaOH

Answer 1

HCl and NaOH are strong acid and strong base respectively so they will completely diassociate in aqueous solution

HCl (aq) -----> H+ (aq) + Cl- (aq)

NaOH ( aq) -----> Na+ (aq) + OH- (aq)

so No. of moles of H+ = No. of moles of HCl

No. of moles of OH- = No. of moles of NaOH

Given

HCl

Volume V1 = 7.5 ml = 0.0075 L

Molarity M1 = 0.15 M (mol/L)

No. of moles (HCl = H+ )n1 = V1 * M1 = 0.0075 * 0.15 = 0.001125 moles

NaOH

Volume V1 = 19.29 ml = 0.001929 L

Molarity M1 = 0.15 M (mol/L)

No. of moles (NaOH = OH- )n1 = V1 * M1 = 0.01929 * 0.15 = 0.0028935 moles

H+ will be nuetralized by OH- so

No. of moles OH- remaining = 0.0028935 moles - 0.001125 moles = 0.0017685 moles

total volume = 7.5 ml + 19.29 ml = 26.79 ml = 0.02679 L

Molarity of OH- remaining in solution =  0.0017685 moles / 0.02679 L = 0.066 mol/L

pOH = -log ( [OH-]) = - log(0.066) = 1.18

pH = 14 - pOH = 12.82 Answer