Question

A solution of `0.1 M NaZ` has `PH = 8.90`. The `K_a` of `HZ` is.
A. `6.3 xx 10^-11`
B. `6.3 xx 10^-10`
C. `1.6 xx 10^-5`
D. `1.6 xx 10^-6`

Answer 1

Correct Answer - C
`NaZ` is salt of `W_A//S_B`
`:. pH = (1)/(2) (pK_w + pK_a + log C)`
`8.9 xx 2 = 14 + pK_a + log 0.1`
`17.8 = 14 + pK_a - 1`
`pK_a = 4.8`,
`K_a = "anti"log (-4.8) = Antilog (-4 - 0.8 -1)`
=`Anilog (overline (5).2) = 1.585 xx 10^-5 ~~ 1.6 xx 10^-5`.