InterviewSolution
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InterviewSolution
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`100mL` of a buffer solution contains `0.1M` 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. `0.328`B. `0458`C. `4.19`D. None |
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Answer» Correct Answer - A For acidic buffer, `pH = pK_(a) + "log" (0.1)/(0.1)` `pH = pK_(a) = - log (10^(-5)) = 5`. Rule: `ABC` (In acidic buffer `(A)`, on addition of `S_(B)(B)`, the concentration of `W_(A)(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)) = Antilog (1) = 10` Solve for `x:` `x = 0.082 M = (0.082)/(1000) xx 100` `= 0.0082 mol (100 mL)^(-1)` `= 0.0082 xx 40 g(100 mL)^(-1)` `= 0.328 g` |
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