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

Step-by-step explanation:

We are given:

Molarity of ammonia = 0.2 M

`K_b=1.8* 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=([NH_4^+][OH^-])/([NH_3])`

Putting values in above expression, we get:

`1.8* 10^(-5)=(x^2)/(0.2-x)\\\\1.8* 10^(-5)(0.2-x)=x^2\\\\x^2+(1.8* 10^(-5)x)-(0.36* 10^(-5))=0\\\\x=1.88* 10^(-3), 1.9* 10^(-3)`

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

So,
`[OH^-]=x=1.88* 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* 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