Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+ 
⇄ 
Formation constant Kf
Kf = / ( [][] ) = 5.0 × 10¹⁰
Now,
[] = 
; ∝₄ = 0.35
so the equilibrium is;
+ 
⇄ + 4H⁺
Given that; 
= { 1 mol reacts with 1 mol }
so at equilibrium, = = x
∴
+ ⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [][] ) = / ( [][] ) = 5.0 × 10¹⁰
⇒ / ( [][] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
