Consider a solution having [H+]=10−5M at 25∘C, then OH− ion concentration of this solution at the given temperature is :
A
10−5M
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B
10−9M
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C
10−10M
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D
10−8M
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Solution
The correct option is B10−9M pH=−log[H+]pH=−log(10−5)pH=5 ∵pH+pOH=14 ∴pOH=14−5=9 −log[OH−]=9 [OH−]=10−9
Theory: pKw: pKw at 25oC is pKw=−log10(1.0×10−14)=14 Since pH=−log10[H+] and pOH=−log10[OH−] Kw=[H+][OH−]−log10Kw=(−log10[H+])+(−log10[OH−]) pKw=pH+pOH, 14=pH+pOH