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0.1 M HQ acid has pH=3 then findits ionization constant. |
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Answer» `3xx10^(-1)` pH=3 `THEREFORE [H^+]=10^(-3)` `therefore x=10^(-3)` `therefore K_a=((x)xx(x))/(0.1-x)` `=(10^(-3))^2/(0.1-10^(-3)) approx 10^(-6)/0.1 = 1xx10^(-5)` |
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