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On adding `0.04g` solid `NaOH` to a `100mL,(M)/(200)Ba(OH)_(2)` solution, determine change in `pH`:A. `0`B. `+0.3`C. `-0.3`D. `+0.7` |
Answer» initial `rArr[OH^(-)]=(2)/(200)=10^(-2)M:.pH=12` final `rArr [OH^(-)]=10^(-2)+(0.04)/(40xx0.1)=2xx10^(-2):.pH=12.3` So change =`12.3-12=+0.3` |
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