Question

Suppose 9.75 mL of 0.010000 M EDTA is required to react with a 50.00 mL sample of saturated MgCO3 solution. Calculate the K. A) 2.51 x 10^-3 B)1.00 x 10^-5 C)9.75 x 10^-3 D)3.80 x 10^-6 E)4.88 x 10^-2

Answer 1

Reaction between Mg^2+ and EDTA is given by

Mg^2+ + EDTA ----> Mg(EDTA)

1mole Mg^2+ 1mole EDTA

Thus, M1V1 = M2V2

Given that, M1 = 0.01 M, V1 = 9.75mL, M2 = ? and V2 = 50mL

M2 = M1V1/V2 = 0.01*9.75/50 = 0.00195 M

Molarity of Mg^2+ ion in 50mL of solution = 0.00195 M

The solubility of MgCO3 is given by,

MgCO3(s) ==== Mg^2+ (aq) + CO3^2-(Aq)

0.00195M ------- 0.00195M + 0.00195M

The equilibrium constant is given by

K = [Mg^2+][CO3^2-] = (0.00195)^2 = 3.80*10^-6

Answer D