Answer to Question #168998 in Organic Chemistry for marcus mendoza

Question

1. Calculate the molar solubility of Ag ion in AgCl? (Ksp = 1.6x10^-10)

AgCl (s) ↔ Ag⁺ (aq) + Cl^- (aq)

2.Ka for acetic acid is 1.75x10^-5. Find Kb for acetate ion at 25^oC. (K_w= 1.0 x 10^-14)

Answer 1

1.We know that AgCl is aqueous solution dissociate as and (Ksp = 1.6x10^-10

AgCl (s) ↔ Ag⁺ (aq) + Cl^- (aq)

take molar solubility as S and it dissociate into two ions as Ag⁺ and Cl^-

$K_{sp}= S\times S$

$S^2= K_{sp}$

$S^2=1.6 \times 10^{-10}$

$S=1.26\times10^{-5}$

2 In the question it is given that

Ka for acetic acid is 1.75x10^-5 and also K_w= 1.0 x 10^-14

we know that

$K_a \times K_b= K_w$

$1.75\times 10^{-5} \times K_b= 1.0 \times 10^{-14}$

$K_b=5.7 \times 10^{-8}$