Question

Please help! Will rate.

Cd" forms two complexes with acetate, with the given overall formation constants. 1. Cd2+ (aq) + CH, COŽ (aq) =C(CH.CO2)+(aq) 2. Cd²+ (aq) + 2CH,CO2 (aq) = CaCH,CO2),(aq) R = B = 85 B2 = 1400 Find the stepwise formation constant, K2, for the reaction 3. Cd(CH, CO)* (aq) +CHCO, (aq) = Ca(CH, CO ) (aq) Kg =? You prepare 1.00L of solution containing 1.00 x 10-mol Ca(CIO), and 0.260 mol CH, CO, Na. Calculate the fraction of cadmium in the form Cd+

Answer 1

Cd2+ + CH3COO- <====> Cd(CH3COO)+

b1 = [Cd(CH3COO)+] / [Cd2+] [CH3COO-] = 85

Cd2+ + 2CH3COO- <====> Cd(CH3COO)2

b2 = [Cd(CH3COO)2] / [Cd2+] [CH3COO-]^2 =1400

Cd(CH3COO)+ + CH3COO- <====> Cd(CH3COO)2

K2 = [Cd(CH3COO)2] / [Cd(CH3COO)+] [CH3COO-]

so K2 = b2 * 1/b1 = 1400/85 = 16

Cd2+ + CH3COO- <====> Cd(CH3COO)+

Cd2+ + 2CH3COO- <====> Cd(CH3COO)2

b1=85

b2 =1400

$\alpha$ = [Cd2+] /[Cd2+] [Cd (CH3COO)+] + [Cd(CH3COO)2] = [Cd2+] / [Cd2+] [ 1+ b1(CH3COO-) +b2(CH3COO-)^2]

$\alpha$ = 0.0312

weight of MnSO4 = 0.535 g

moles of MnSO4 = 0.535/151 = 3.54*10^-3 mol

molarity of solution = 6.81*10^-3 M

strength of EDTA = 0.046*6.81*10^-3/0.0415 = 7.55*10^-3 M

moles in 1.70 ml of this EDTA solution = 1.28*10^-5 mol

so moles of CaCO3 = 1.28*10^-5 mole

weight of CaCO3 = 1.28*10^-5*100.09 = 0.00128 g = 1.28 mg