Question

The conditional formation constant for the calcium -EDTA complex at pH 10 is 1.80x10^10. Consider the titration of 50.00ml of 0.0100 M Ca^2+ with a solution of 0.0100 M EDTA at pH 10. Calculate pCa after 10.000ml of titrant has been added. Enter your answer below, to a precision of two digits after the decimal point.

Answer 1

Given pH of the solution is 10.0 and the conditional formation constant, K_f’ = 1.8*10¹⁰.

Moles Ca²⁺ initially present = (50.00 mL)*(1 L/1000 mL)*(0.0100 mol/L) = 5.0*10^-4 mol.

Moles EDTA added = (10.00 mL)*(1 L/1000 mL)*(0.0100 mol/L) = 1.0*10^-4 mol.

Consider the reaction between Ca²⁺ and EDTA as

Ca²⁺ + EDTA ------> CaEDTA^2-

Moles Ca²⁺ reacted = moles EDTA added = 1.0*10^-4 mol.

Therefore, moles Ca²⁺ remaining = (5.0*10^-4 – 1.0*10^-4) mol = 4.0*10^-4 mol.

Total volume of solution = (50.00 + 10.00) mL = 60.00 mL.

Therefore, [Ca²⁺] remaining = (4.0*10^-4 mol)/[(60 mL)*(1 L/1000 mL)] = 6.667*10^-3 mol/L

pCa²⁺ = -log [Ca²⁺]_remaining = -log (6.667*10^-3) = 2.176 ≈ 2.18 (ans).