JEE MAIN - Chemistry (2021 - 25th July Evening Shift - No. 21)
Assuming that Ba(OH)2 is completely ionised in aqueous solution under the given conditions the concentration of H3O+ ions in 0.005 M aqueous solution of Ba(OH)2 at 298 K is ______________ $$\times$$ 10$$-$$12 mol L$$-$$1. (Nearest integer)
Answer
1
Explanation
$$Ba{(OH)_2} \to B{a^{ + 2}} + 2O{H^ - } \downarrow 2 \times 0.005 = 0.01 = {10^{ - 2}}$$
At 298 K : in aq. solution $$[{H_3}{O^ + }][O{H^ - }] = {10^{ - 14}}$$
$$[{H_3}{O^ + }] = {{{{10}^{ - 14}}} \over {{{10}^{ - 2}}}} = {10^{ - 12}}$$
At 298 K : in aq. solution $$[{H_3}{O^ + }][O{H^ - }] = {10^{ - 14}}$$
$$[{H_3}{O^ + }] = {{{{10}^{ - 14}}} \over {{{10}^{ - 2}}}} = {10^{ - 12}}$$
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