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Ionisation constant of `CH_(3)COOH` is `1.7xx10^(-5)` and concentration of `H^(+)ions` is `3.4xx10^(-4)`. Then, find out initial concentration of `CH_(3)COOH` molecules.A. `3.4xx10^(-4)`B. `3.4xx10^(-3)`C. `6.8xx10^(-4)`D. `6.8xx10^(-3)` |
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Answer» Correct Answer - D `CH_(3)COOH hArr CH_(3)COO^(-)+H^(+)` `K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])` Given that, `[CH_(3)COO^(-)]=[H^(+)]=3.4xx10^(-4)M` `K_(a)` for `CH_(3)COOH=1.7xx10^(-5)` `CH_(3)COOH` is weak acid, so in it `[CH_(3)COOH]` is equal to initial concentration. Hence, `1.7xx10^(-5)=((3.4xx10^(-4))(3.4xx10^(-4)))/([CH_(3)COOH])` `[CH_(3)COOH]=(3.4xx10^(-4)xx3.4xx10^(-4))/(1.7xx10^(-5))` `=6.8xx10^(-3)M` |
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