Question

By evaporating the solvent from the extract, she determines that there is only 0.124 g of acid per 10.0 mL of plant extract. What is the ka value for the acid? What is the pH of the plant extract? What is the pH after the following information, calculate the solubility product of AgCN. Calculate the solubility if the pH = 4.75 and the [HCN] is 0.130 M, and the pK_a for HCN is 9.31.

Answer 1

pH = pKa + log[salt]/[acid]

given pH= 4.75, pka= 9.31 [HCN]= 0.130

4.75 = 9.31 + log [salt]/0.130

log [salt]= -5.446, [salt] = 10^-5.446 = 3.5806 x 10^-6

hence [AgCN]= 3.5806 x 10^-6

hence Solubility, S = 3.5806 x 10^-6

$\Delta Go= -2.303RT log K$

$\Delta Go= -nFEocell$

Eo cell= E(cathode)- E(anode) = 0.80 - (-0.07) = 0.87v

$\Delta Go=$ - 1 * 96500 * 0.87 = - 83955 J

$\Delta Go= -2.303RT log K$

-83955 = -2.303 (8.314) (298) log K

log k= 14.713, K = 5.16 x 10¹⁴

dissociation of AgCN= Ag⁺ + CN^-

relation betwen Ksp and solubility

Ksp = S^(x+y) * x^x * y^y

Coefficient of Ag (x)=1

Coefficient of CN(y)=1

Ksp = (3.5806 x 10^-6)² = 7.1612 x 10 ^-12