Question

calculate the pH of a 0.2 M * 4 solution for which Kb = 1.8*10^-5 at 26 c . The equation for the reaction
Nh3+H2O->NH4+oh

Answer 1

Answer: The pH of the solution is 11.24

Explanation:

We are given:

Molarity of ammonia = 0.2 M

$K_b=1.8\times 10^{-5}$

The given chemical equation follows:

                $NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

I:                0.2

C:               -x                      +x        +x

E:             0.2-x                     x          x

The expression for equilibrium constant follows:

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Putting values in above expression, we get:

$1.8\times 10^{-5}=\frac{x^2}{0.2-x}\\\\1.8\times 10^{-5}(0.2-x)=x^2\\\\x^2+(1.8\times 10^{-5}x)-(0.36\times 10^{-5})=0\\\\x=1.88\times 10^{-3}, 1.9\times 10^{-3}$

Neglecting the negative value of x as concentration cannot be negative.

So, $[OH^-]=x=1.88\times 10^{-3}M$

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.

$pOH=-\log [OH^-]$

Putting values in above equation, we get:

$pOH=-\log (1.88\times 10^{-3})\\\\pOH=2.76$

We know:

$pH+pOH=14\\\\pH=14-2.76\\\\pH=11.24$

Hence, the pH of the solution is 11.24