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)

Answer»

`7.00`
7.87
5.13
6.13

Solution :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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