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Question

$${K}_{w}$$ for $${H}_{2}O$$ is $$9.62\times {10}^{-14}$$ at $${60}^{o}C$$. What is $$pH$$ of water at $${60}^{o}C$$.


A
3.71
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B
6.51
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C
7.49
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D
11.31
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Solution

The correct option is C $$6.51$$
We know that when $$K_{w}=10^{-14}$$, 
$$[H_{3}O^{+}]=[OH^{-}]=10^{-7} mol/dm^{3}$$
The value of $$K_{w}$$ depends on temperature. So, in the question, the value is not $$10^{-14}$$.
SImilarly, we can use the relation for above given value of $$K_{w}$$
So,
$$[H_{3}O^{+}]= \sqrt{9.62*10^{-14}}=3.1\times 10^{-7} mol/dm^{3}$$ 
Hence,
$$pH=-log_{10}[H_{3}O^{+}]=-log_{10}(3.1*10^{-7})=7-0.49=6.51$$
Shortcut:
Remember that you cannot use log tables in your exam. So, here once you find the value of $$[H_{3}O^{+}]$$, taking log will give you "7 - something lesser than 1" which matches with one option only i.e. B.

Chemistry

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