InterviewSolution
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InterviewSolution
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`100 mL` of a buffer solution contains `0.1 M` each of weak acid `HA` and salt `NaA`. How many gram of `NaOH` should be added to the buffer so that it `pH` will be `6` ? (`K_(a)` of `HA=10^(-5)`).A. `4.19`B. `0458`C. `0.328`D. None |
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Answer» Correct Answer - C For acidic buffer, `pH=pK_(a)+log``(0.1)/(0.1)` `pH=pK_(a)= -log(10^(-5))=5`. Rule: `ABA` (In acidic buffer (`A`), on addition of `SB(B)`, the concentration of `WA(A)` decreases and that of salt increases). Let `x M` of `NaOH` is added. `pH_(new)=5+log``((0.1+x)/(0.1-x))` `6-5= log``((0.1+x)/(0.1-x))` `((0.1+x)/(0.1-x)) = anti log (1)=10` Solve for `x`: `x=0.082 M=(0.082)/(1000)xx100` `=0.0082 mol (100 mL)^(-1)` `=0.0082xx40 g (100mL)^(-1)` `0.328 g` |
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