Question

Calculate pCa2+ for the titration of 25.00 mL of 0.0200 M EDTA with 12.0 mL of 0.01000 M CaSO4 at pH 10.00.

Answer 1

Molarity Of EDTA = 0.0200 M

Volume of EDTA = 25 mL = 0.025 L

moles of EDTA = Molarity * volume = 0.0200 M * 0.025 L = 0.00050 mol

Molarity of Ca²⁺ = 0.0100 M

Volume of Ca²⁺ = 12 mL = 0.012 L

Moles of Ca²⁺ = 0.01 M * 0.012 L = 0.000120 moles

1 mol of EDTA reacts with 1 mol of Ca²⁺

Thus, 0.00012 mol of Ca²⁺ reacts with 0.000120 mol of EDTA

Thus, Moles of EDTA left = 0.00050-0.00012 = 0.000380 moles

Total volume of solution = 25 mL + 12 mL = 37 mL = 0.037 L

Thus, [EDTA] = Moles/ volume = 0.000380 moles/ 0.037L = 0.0103 M

Now moles of Ca(EDTA)^2- forms = moles of Ca²⁺ reacts = 0.00012 moles

So, [Ca(EDTA)²⁺] = moles/ Total volume = 0.00012moles/ 0.037 L = 0.00324 M

ICE table is:

Ca²⁺ + EDTA Ca(EDTA)²⁺

I(M) 0 0.0103 0.00324

C(M) +x +x -x

E(M) x 0.0103+x 0.00324 -x

so, K_f = [Ca(EDTA)²⁺]/ ([Ca²⁺ ][EDTA] )

From literature, K_b = 1.76 * 10¹⁰

1.76 * 10¹⁰ = (0.00324-x)/( (x)(0.0103-x))

But 0.00324-x 0.00324

Also . 0.0103+x 0.0103

1.76 * 10¹⁰ = 0.00324/ 0.0103x

x = 1.79 * 10^-11 = [Ca²⁺]

So, pCa²⁺ = -log [Ca²⁺] = -log (1.79 * 10^-11 ) = 10.75