Question

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.

Answer 1

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

$Ca^{2+$ + $y^{4-$ ⇄  $CaY^{2-$

Formation constant Kf

Kf = $CaY^{2-$ / ( [$Ca^{2+$][$y^{4-$] ) = 5.0 × 10¹⁰

Now,

[$y^{4-$] = $\alpha _4CH_4Y$; ∝₄ = 0.35

so the equilibrium is;

$Ca^{2+$ + $H_4Y$ ⇄  $CaY^{2-$ + 4H⁺

Given that; $CH_4Y$ = $Ca^{2+$     { 1 mol $Ca^{2+$  reacts with 1 mol $H_4Y$  }

so at equilibrium, $CH_4Y$ = $Ca^{2+$ = x

∴

$Ca^{2+$ + $y^{4-$ ⇄  $CaY^{2-$

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = $CaY^{2-$ / ( [$Ca^{2+$][$y^{4-$] ) =  $CaY^{2-$ / ( [$Ca^{2+$][$\alpha _4CH_4Y$] )  = 5.0 × 10¹⁰

⇒ $CaY^{2-$ / ( [$Ca^{2+$][$\alpha _4CH_4Y$] )  = 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