`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))`.

Options

A. `0.328`
B. `0458`
C. `4.19`
D. None

Answer

A

Explanation

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`