pKa.pH HO2 - ETX 102A Influence of pH on Water Solubility 1...

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1 ETX 102A Influence of pH on Water Solubility 1. Weak Acid Benzoic Acid (p K a = 4.0), equation: pH = p K a + log [A - ]/[HA] Solve for [A - ]/[HA] a. HA versus A - in freshwater, pH 4.0: pH p K a = log [A - ]/[HA] 4.0 4.0 = log [A - ]/[HA] 0 = log [A - ]/[HA] antilog 0 = [A - ]/[HA] 1 = [A - ]/[HA] (or a 1:1 ratio; note the total mass is 1 + 1 = 2) A - = 1/2 x 100 = 50% and HA = 1/2 x 100 = 50% b. HA versus A - in seawater, pH 8.0: pH p K a = log [A - ]/[HA] 8.0 4.0 = log [A - ]/[HA] 4.0 = log [A - ]/[HA] antilog 4.0 = [A - ]/[HA] 10,000 = [A - ]/[HA] (or a 10,000:1 ratio; note the total mass is 10,000 + 1 = 10,001) A - = 10,000/10,001 x 100 = 99.99% and HA = 1/10,001 x 100 = 0.01% Conclusion Benzoic acid is more soluble in seawater. 2. Weak Base Aniline (p
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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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