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If Ka of weak acid is found to be 1.78xx...

If Ka of weak acid is found to be `1.78xx10^(-5)` What is the `PK_(a)` value?

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`Ka=1.78xx10^(-5)`
`PK_(a)=-logKa=-log(1.78xx10^(-5))=-log1.78+5log10=-0.2504+5=4.7496`
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