1.

At `100^(@)C` the `K_(w)` of water is 55 times its value at `25^(@)C`. What will be the pH of neutral solution (log 55 = 1.74)A. `7.00`B. 7.87C. 5.13D. 6.13

Answer» Correct Answer - D
At `100^(@)C`
`K_(w) = 55 xx 10^(-14)`
`H^(+) = sqrt(55 xx 10^(-14))`
`= 7.41 xx 10^(-7)`
`pH = -log [H^(+)]`
`= -log[7.41 xx 10^(-7)]`
`= -[log 7.41 + log 10^(-7)]`
`= -[0.86 - 7] = -[-6.13] = 6.13`


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