Question

19. A. Solid calcium nitrate is slowly added to 175 mL of a 0.0676 M potassium carbonate solution. The concentration of calcium ion required to just initiate precipitation is _____ M.

B. Solid zinc acetate is slowly added to 125 mL of a 0.0515 M potassium cyanide solution. The concentration of zinc ion required to just initiate precipitation is _____ M.

C. Solid sodium hydroxide is slowly added to 150 mL of a 0.0435 M copper(II) acetate solution. The concentration of hydroxide ion required to just initiate precipitation is _____ M.

Answer 1

A)

Let us consider a reaction,

CaNO₃ + K2CO3 --> CaCO3 + K2NO3

We know ,Ksp of CaSO4 = 3.36 *10^-9

CaCO3 --> Ca²⁺ + CO₃^2-

[CO₃^2-] = 0.0676 M

We know,

Ksp = [Ca2+] [CO₃^2-]

  3.36 *10^-9 = [Ca₂⁺] x 0.0676

[Ca₂⁺] = 4.97 x 10^-8M

B)

Let us consider a reaction,

(CH3COO)₂Zn + 2KCN -----> 2CH3COOK + Zn(CN)₂

From the reaction , 2 moles of KCN are required against each ion of Zn.

Number of moles of KCN are = 0.0515 x 0.125

= 0.00643 moles

So , the number of moles of Zn ion = 0.00643 / 2

= 0.003215 moles

Concentration in 125 mL will be = 0.003215 / 0.125

= 0.02572 M

C)

Let us consider a reaction,

Cu(OH)2(s) <--------> Cu2+(aq) + 2OH- (aq)

Ksp of Cu(OH)2 = 2.20×10^-20

Concentration of Copper(II) acetate = 0.0435M

Concentration of Cu2+ = 0.0435M

Therefore, Ksp = [ Cu2+ ] [ OH- ] ^2

2.20× 10^-20 = 0.0435 × [ OH- ]^2

[ OH- ]^2 = 2.20× 10^-20 / 0.0435

= 5.05 × 10^-19

[ OH-] = 2.24×10^-9M

So, Concentration of OH- ion required = 2.24×10^-9M