Question

A 250 mL sample of saturated AgOH solution was titrated with HCl, and the endpoint was reached after 2.60 mL of 0.0136 M of HCl was dispensed. Based on this titration, what is the Ksp of AgOH?

Answer 1

Answer:

Ksp=1.9×10⁻⁸

Explanation:

First write the balane chemical reaction

AgOH + HCl → AgCl+H2O

Titration reaction occur completely

N-factor of AgOH and HCl is one

normality=molarity

Noermality of HCl=0.0136

N₁ and N₂ are the normality of AgOH and HCl respectively

V₁ and V₂ are the volume of AgOH and HCl respectively

mili equvalent of AgOH-mili equivalent of HCl

$N_1$=?

$V_1$=250ml

$N_2$=0.0136

$V_2$=2.6ml

$N_1V_1=N_2V_2$

after puttting all the value we get,

$N_1$=0.000141 =normality of AgOH solution

molarity=0.000141M

Calculation of Ksp

AgOH⇄Ag⁺  +  OH⁻

Lets solubility=S

$[Ag^+]$=S

[OH⁻]=S

and S is equal to the concentration of AgOH because that concentration is calculated when solution become satrated.

hence S=0.000141M

Ksp=[Ag+][OH-]

Ksp=S×S

Ksp=1.9×10⁻⁸