1.

The dissociation constant of water is `1 xx 10^(-14) mol ^(-2) litre ^(-2)`. What is the pH of 0.001 M KOH soluiton ?A. `10^(-11)`B. `10^(-3)`C. 3D. 11

Answer» Correct Answer - D
`[OH^(-)]=[KOH]=0.001M =1xx10^(-3)M`
`:. p(OH)=-log (1xx 10^-3))=3`
`pH=14-p(OH)=14-3=11`


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