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Find out the `OH^(-)` ion concentration in `100 mL` of `0.015 M HCl` isA. `2.0 xx 10^(-9) M`B. `6.7 xx 10^(-13) M`C. `3 xx 10^(-10) M`D. `5 xx 10^(-12) M` |
Answer» Correct Answer - B `[OH^(-)] = (Kw)/([H^(+)]` `= (1 xx 10^(-14))/(0.015)` `= 6.7 xx 10^(-13)` |
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