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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 |
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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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