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pH of 0.1 M `NH_(3)` aqueous solution is `(K_(b) = 1.8 xx 10^(-5))`A. 11.13B. 12.5C. 13.42D. 11.55 |
Answer» Correct Answer - A `NH_(4)OH hArr NH_(4)^(+) + OH^(-)` `K_(b) = C alpha^(2) , (1.8 xx 10^(-5))/(.1) = alpha^(2), alpha = 1.34 xx 10^(-3)` `[OH^(-)] = alpha. C = 1.34 xx 10^(-3) xx .1` `pOH = log 10(1)/(1.34 xx 10^(-4)) , pOH = 2.87` `pH = pOH = 14 , pH + 2.87 = 14` `pH = 14 - 2.87, pH = 11.13`. |
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