20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. `K_(a)` of acetic acid = `1.8 xx 10^(-5)` at `27^(circ)C`. Calculate concentration of potassium acetate if pH of the mixture is 4.8.

Options

A. 0.1M
B. 0.04M
C. 0.4M
D. 0.02M

Answer

B

Explanation

`K_(a) = 1.8 xx 10^(-5)`
pH = 4.8
Let the concentration of pottasium acetate = x
`pH = pK_(a) + "log"[sal t]/[acid]`
`4.8 = "log"(1.8 xx 10^(-5))"log" X xx 50/(20 xx 0.1)`
`implies 4.8 = 4.74 + "log" 25x`
`implies "log" 25x = 0.06`
`25x = 1.148 (:' "log" 1.148 = 0.06)`
x = 0.045M