Calculate the pH of `0.1 M NH_4 OH` if
`K_b =1.75 xx10^(-5)`.

Explanation

Degree if dissociation `alpha=sqrt((K_a)/(C )), C alpha = sqrt(K_a . C)`
`therefore pOH=log.(1)/(sqrt(K_b. C))=log.(1)/(sqrt(1.75xx10^(-5)xx1)`
`=log.(1)/(sqrt(1.75xx10^(-6)))=log .(10^3)/(sqrt(1.75))`
`=3-(1)/(2) log 1.75`
`=3-(1)/(2)xx0.2430 =2.8785`
`therefore pH=14-pOH`
`=14-2.8785=11.1215`
Alternating,
`[OH^-]=sqrt(K_bxx C)`
`=sqrt(1.75xx10^(-5)xx0.1)=sqrt(1.75xx10^(-6))`
`=1.322xx10^(-3)`
`therefore pOH=-log [H^+]=-log (1.323xx10^(-3))`
`=3-0.1216=2.8784`
`therefore pH=14-pOH=14-2.8784=11.1216`.