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Question
If an aqueous solution at 25oC, has twice as many OH− as pure water, its pOH will be:
If an aqueous solution at 25oC, has twice as many OH− as pure water, its pOH will be:
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Solution
The correct option is C 6.69
[OH−] for pure water at 25oC=10−7
[OH−] for the aqueous at 25oC=2×10−7
Hence, pOH−log[OH−]=−log(2×10−7)=7−log2=6.698
[OH−] for pure water at 25oC=10−7
[OH−] for the aqueous at 25oC=2×10−7
Hence, pOH−log[OH−]=−log(2×10−7)=7−log2=6.698
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