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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` |
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Answer» 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`. |
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