Question

1. An aqueous solution contains 7.65×10^-2 M KHCO₃ and 0.245 M H₂CO₃.  

The pH of this solution is

2. An aqueous solution contains 0.207 M KHCO₃ and 0.466 M H₂CO₃.  

The pH of this solution is

3. A student measures the SO₃^2- concentration in a saturated aqueous solution of silver sulfite to be 1.52×10^-5 M.  
Based on her data, the solubility product constant for silver sulfite is

4. A student measures the molar solubility of zinc phosphate in a water solution to be 1.51×10^-7 M.  
Based on her data, the solubility product constant for this compound is

Answer 1

1) Solution of KHCO₃ and H₂CO₃ is a Buffer solution and it's pH is calculated by using Henderson's equation

pH = pKa₁ + log [ Salt ] / [ Acid]

= 6.37 + log [KHCO₃ ] / [H₂CO₃ ]

= 6.37 + log ( 7.65 x 10 ^-2) / 0.245

= 6.37 + log 0.312

= 6.37 - 0.506

= 5.86

2)

pH = pKa₁ + log [ Salt ] / [ Acid]

= 6.37 + log [KHCO₃ ] / [H₂CO₃ ]

= 6.37 + log ( 0.207) / 0.466

= 6.37 + log 0.444

= 6.37 - 0.352

= 6.02

3) Consider dissociation of Silver sulphite.

Ag₂(SO₃)_(s) 2 Ag ⁺_(aq) + SO₃ ^2-_(aq)

[ SO₃ ^2- ] = 1.52 x 10 ^-05 M then [ Ag ⁺ ] = 2 x 1.52 x 10 ^-05 M = 3.04 x 10 ^-05 M

We have , K sp = [Ag ⁺ ] ² [ SO₃ ^2-]  

= ( 3.04 x 10 ^-05 ) ² x 1.52 x 10 ^-05

= 1.40 x 10 ^-14  

4)

Consider dissociation of Zinc phosphate.

Zn₃(PO₄)_2(s) 3 Zn ²⁺_(aq) +2 PO₄ ^3-_(aq)

Ksp = [Zn ²⁺ ] ³ [ PO₄ ^3- ] ²

If S = Solubility of Zn₃(PO₄)₂ in mol / L , then [ Zn ²⁺ ]  = 3 S and [PO₄ ^3-] = 2 S

Therefore,   [ Zn ²⁺ ] = 3 x 1.51 x 10 ^-07 = 4.53 x 10 ^-07 mol / L

[PO₄ ^3-] = 2 x 1.51 x 10 ^-07 =  3.02 x 10 ^-07 mol / L

Hence, Ksp = ( 4.53 x 10 ^-07) ³ x (  3.02 x 10 ^-07) ²

= 8.48 x 10 ^{- 33}