Unformatted text preview: K a = 5.0), equation for weak bases: pH = p K a + log [B]/[BH + ] Solve for [B]/[BH + ] a. BH + versus B in freshwater, pH 4.0: pH – p K a = log [B]/[BH + ] 4.0 – 5.0 = log [B]/[BH + ] 1.0 = log [B]/[BH + ] antilog 1.0 = [B]/[BH + ] 0.1 = [B]/[BH + ] (or a 1:10 ratio; note the total mass is 10 + 1 = 11) B = 1/11 x 100 = 9.1% and BH + = 10/11 x 100 = 90.9% b. BH + versus B in seawater, pH 8.0: pH – p K a = log [B]/[BH + ] 8.0 – 5.0 = log [B]/[BH + ] 3.0 = log [B]/[BH + ] antilog 3.0 = [B]/[BH + ] 1000 = [B]/[BH + ] (or a 1000:1 ratio; note the total mass is 1000 + 1 = 1001) B = 1000/1001 x 100 = 99.9% and BH + = 1/1001 x 100 = 0.1% Conclusion – Aniline is more soluble in freshwater....
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 Winter '10
 RonaldTjeerdema
 pH, 50%, 9.1%, 0.1%, 0.01%, 90.9%

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