Calculate volume (liters) of 1.572 M KOH that must be added to a 0.112 L solution of 9.80 g of (H3Glu Cl–, MW = 183.59 g/mol) for pH 10.35?

Consider the third dissociation of glutamic acid :
HGlu²⁻(aq) ⇌ H⁺(aq) + Glu³⁻ …. Ka₃ = 9.95
The pKa is close to the pH required (10.35).

n(H₃Glu) in the solution = (9.80 g) / (183.59 g/mol) = 0.05338 mol

H₃Glu(aq) + 2OH⁻(aq) → HGlu²⁻(aq) + 2H₂O(l)
When 0.05338 × 2 mol of KOH is added, all 0.05338 mol H₃Glu is converted to 0.05338 mol HGlu²⁻.

Let y mol be the no. of moles of further KOH solution added to make the pH 10.35.
HGlu²⁻(aq) + OH⁻(aq) → Glu³⁻(aq) + H₂O(l)
n(Glu³⁻) in the final solution = y mol
n(HGlu²⁻) in the final solution = (0.05338 - y) mol

According to the above chemical equation :
pH = pKa₃ + log{[Glu³⁻]/[HGLu²⁻]}
pH = pKa₃ + log{n(Glu³⁻)/n(HGLu²⁻)}
10.35 = 9.95 + log{y/(0.05338 - y)}
log{y/(0.05338 - y)} = 0.4
y/(0.05338 - y) = 10⁰·⁴
y/(0.05338 - y) = 2.512
y = 0.1341 - 2.512y
3.512y = 0.1341
y = 0.03818

Total n(KOH) added = (0.05338 × 2 + 0.03818) mol = 0.1449 mol
Volume of KOH solution added = (0.1449 mol) / (1.572 mol/L) = 0.0922 L