Question

1) A saturated solution of copper (II) iodate in pure water has a copper ion concentration of 2.7X10^-3 M.

a) whar is the molar solubility of copper iodate in a 0.35 M Potassium iodate solution

B) what is the molar solubility of copper Iodate in 0.35 M copper nitrate solution

2)

a)consider the slightly soluble salt, silver chloride, Agcl. If you had 10.0 g of silver chloride, what volume of water would be required to completely dissolve this amount

b) what volume of 15 M NH3 would be required to dissolve the same quantity of AgCl completey

Answer 1

1 Ans.

The dissociation of Cu(IO₃)₂ is

$Cu(IO_{3})_{2}\rightleftharpoons Cu^{2+}+ 2IO_{3}^{-}$

The K_sp expression is

where x is the molar solubility of Cu(IO₃)₂ and also the concentration of copper ion. The solution is a saturated solution. So, the concentration of Cu²⁺ is the same as the molar solubility.

Given that,

$x = 2.7 \times 10^{-3}M$

Therefore,

$K_{sp} = 4\times (2.7\times 10^{-3})^{3}$

$K_{sp} = 7.87 \times 10^{-8}$

Therefore, K_sp of Cu(IO₃)₂ is 7.87 x 10^-8

a Ans.

Molarity of KIO₃ = 0.35M

[IO₃^-] from KIO₃ = 0.35M

K_sp of Cu(IO₃)₂ as written before is

But, 0.35 has to be added to the concentration of IO₃^-

Therefore,

The value of 2x is very negligibly small compared to 0.35M. So, 2x is neglected to simplify the calculation. Neglecting the 2x term will cause little or no difference.

Substitute the value of K_sp

$7.87\times 10^{-8} = x(0.35)^{2}$

$7.87\times 10^{-8} = 0.1225x$

$\frac{7.87\times 10^{-8} }{0.1225}= x$

$x = 6.4 \times 10^{-7} mol/L$

The molar solubility of copper iodate in 0.35M potassium iodate solution is 6.4 x 10^-7mol/L.

b Ans.

Concentration of Cu²⁺ from copper nitrtate solution = 0.35M

K_sp of Cu(IO₃)₂ as written before is

To the concentration of copper ion, 0.35M has to be added which is already present in the copper nitrate solution.

Therefore,

x, being very small compared to 0.35 is neglected to simplify calculation.

$K_{sp} = 0.35\times 4x^{2}$

$7.87\times 10^{-8} = 1.40x^{2}$

$\frac{7.87\times 10^{-8}}{1.40}= x^{2}$

$\sqrt{\frac{7.87\times 10^{-8}}{1.40}}= x$

$x = 2.4 \times 10^{-4}mol/L$

Therefore, the molar solubility of copper iodate in 0.35M copper nitrate solution is 2.4 x 10^-4 mol/L.

2 a Ans

The K_sp value of AgCl at 25 degree celcius = 1.8 x 10^-10

$AgCl\rightleftharpoons Ag^{+}+Cl^{-}$

$1.8 \times 10^{-10} =x^{2}$

$\sqrt{1.8 \times 10^{-10} }=x$

$x= 1.34 \times 10^{-5}mol/L$

Molar mass of AgCl = 143.32g/mol

143.32g of AgCl is 1 mol.

Therefore, 10.0g of AgCl is

$= \frac{1}{143.32g/mol}\times 10.0g$

$Molar solubility, x= 1.34 \times 10^{-5}mol/L$

1.34 x 10^-5 mol of AgCl is soluble in 1 L

So, 0.0698 mol of AgCl will be soluble in

$= \frac{1}{1.34\times 10^{-5} mol/L}\times 0.06977mol$

Therefore, 5207L of water is required to completely dissolve 10.0g AgCl.

b Ans. Molarity of NH₃ = 15M

$AgCl(s)\rightleftharpoons Ag^{+}(aq)+ Cl^{-}(aq)$

K_sp = 1.8 x 10^-10

$2NH_{3}(aq)+ Ag^{+}(aq)\rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)$

K_f = 1.6 x 10⁷

The overall reaction is

$AgCl(s)+2NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)+ Cl^{-}(aq)$

The equilibrium constant of the overall reaction is

$K = K_{sp}\times K_{f}$

$K = 1.8 \times 10^{-10} \times 1.6 \times 10^{7}$

$K = 2.88 \times 10^{-3}$

The ICE table for the overall reaction is

$AgCl(s)+2NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)+ Cl^{-}(aq)$

	AgCl(s)	NH3(aq)	[Ag(NH3)2]+	Cl-
I	-	15M	0	0
C	-	-2x	+x	+x
E	-	15-2x	x	x

$K = \frac{[Ag(NH_{3})_{2}]^{+}][Cl^{-}]}{[NH_{3}]^{2}}$

$2.88 \times 10^{-3}= \frac{x^{2}}{(15-2x)^{2}}$

Take square root on both sides,

$\sqrt{2.88 \times 10^{-3}}= \frac{x}{15-2x}$

$0.05367= \frac{x}{15-2x}$

$x =\frac{0.8051}{1.1073}$

0.7271 mol of AgCl is soluble in 1L ammonia

10.0g AgCl has 0.0698 mol (calculated in part a)

0.0698 mol will be soluble in

$=\frac{1}{0.7271 mol/L}\times 0.0698mol$

$=0.096L$

$=96mL$

Therefore, 0.096L of 15M ammonia is required.

(N.B: Substitute the exact values of the constants given to you to get the best result and follow the same approach to solve.)