Solubility Products and Water Ionisation Constant

Question 1

What is solubility product

Answer 1

The equilibrium constant for a solubility equilibrium (K_sp)

Question 2

If K= ([Ag⁺(aq)] x [Cl^-(aq]) / [AgCl(s)] how do you find the K_sp

Answer 2

Ksp = [Ag⁺] x [Cl^-]

Question 3

Why is the bottom of the K equation ignored when calculating the K_sp value

Answer 3

The molar concentration of a pure solid is a constant and is independent of the amount present. So it is always equal to 1

Question 4

Why is water ignored in the K_sp equation

Answer 4

The concentration does not change during the reaction

Question 5

Calculate the K_sp from the the following solubility measurements:

PbCl₂(s) <=> Pb²⁺(aq) + 2Cl^-(aq)

Pb concentration in a saturated solution= 1.62x10^-2 moldm^-3

Answer 5

K_sp= [Pb²⁺] x [Cl^-]²

= [1.62x10^-2] x [(1.62x10^-2)x 2]²

= 1.7x10^-5

Question 6

Calculate solubility from K_sp.

K_sp= 5x10^-9

What is the solubility of CaCO₃^2- in moldm^-3

Answer 6

CaCO₃(s) <=> Ca²⁺(aq)+ CaCO₃^2-(aq)

5x10^-9= [Ca²⁺] x [CaO₃^-2]

Since it is a 1:1 ratio of products

5x10^-9= [Ca²⁺]²

so [Ca²⁺] = √5x10^-9

[Ca²⁺] = 7.06x10^-5 moldm^-3

So [CaCO₃] = 7.06x10^-5 moldm^-3

Solulbility = 7.1x10^-5 moldm^-3

Question 7

How do you work out K_w (auto-ionisation of water)

Answer 7

2H₂O(l) <=> H₃O⁺(aq) + OH^-(aq)

K = ([H₃O⁺] x [OH^-]) / [H₂O]²

K_w= [H₃O⁺] x [OH^-]

K_w = 1x10^-14

Question 8

What is the value of K_w

Answer 8

1x10^-14

Question 9

K_w is a fixed value, what affect does this have on equilibrium

Answer 9

Equilibrium shifts to ensure that K_w remains the same. So if [OH^-] increases then [H₃O⁺] decreases