`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)`).

Options

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

Answer

C

Explanation

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`