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 DeltaA= 10^(-5))`

Options

A. 0.328
B. 0.458
C. 4.19
D. 1.32

Answer

A

Explanation

For acidic buffer `pH = pK_(a) + log""(0.1)/(0.1), pH = pK_(a) - log(10^(-5)) = 5`.
Rulee : ABA (In acidic buffer (A), on addition of `S_(B)` (B), the concentration of `W_(A)` (A) decreases and 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 = 0.082` mol `(100 mL)^(-1) = 0.0082 xx 40g (100 mL)^(-1) = 0.328 g`