Question

Calculate the `pH` of a solution obtained by mixing equal volume of `0.02MHOCl &0.2MCH_(3)COOH` solutions
Given that `K_(a)(HOCl)=2xx10^(-4),K_(a)(CH_(3)COOH)=2xx10^(-5)`
Also calculate `[OH^(-)],[OCl^(-)],[CH_(3)COOH]` at equilibrium .Take `log2=0.3`

Answer 1

Volume of final solutions becomes double .
So , concentrated becomes half so after mixing:
`C_(1)=0.01M,C_(2)=0.1M`
`[H^(+)]=sqrt(C_(1)K_(a_(1))+C_(2)K_(a_(2)))=sqrt(2xx10^(-4)xx0.01+2xx10^(-5)xx0.1)=sqrt(2xx10^(-6)+2xx10^(-6))=2xx10^(-3)M`
`:.pH=3-log2=2.7`
`[OCl^(-)]=(0.01xx2xx10^(-4))/(2xx10^(-3))=1xx10^(-3)M,[CH_(3)COO^(-)]=(0.1xx2xx10^(-4))/(2xx10^(-3))=1xx10^(-3)M`,
`[OH^(-)]=(K_(w))/([H^(+)]=(10^(-14))/(2xx10^(-3))=5xx10^(-12)M`