Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for CaY2- is 5.0 x 1010 and the fraction of unprotonated EDTA (Y4-) is 0.35 at pH10. Calculate the concentration of free Ca2 in this solution.

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

just did the assignment and the answer was #1 +2 n