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The pH of `1 M PO_(4)^(3-) (aq)` solution is, [ Given `pK_(b)=(PO_(4)^(3-))=1.62]`A. 13.19B. 1.62C. 8.1D. 4.86 |
Answer» Correct Answer - A `PO_(4)^(3-)+H_(2)O hArr HPO_(4)^(2-)+OH^(-)` `[OH^(-)]=sqrt(K_(b)C)` `pOH =(1)/(2)[pK_(b)-log C]` `=(1)/(2)[1.62-log 1 ] =0.81` `pH =14-pOH =14-0.81=13.19` |
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