# s_monoprotic - Monoprotic acid-base equilibria Weak acid...

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1 Chem215/P.Li/Monoprotic Acid-Base Equilibria/P 1 Monoprotic acid-base equilibria (Ch 10) Weak acid equilibria Fraction of dissociation of a weak acid Weak base equilibria Fraction of association of a weak base When the dissociation of water cannot be neglected Buffer, Henderson-Hasselbalch equation and buffer capacity Chem215/P.Li/Monoprotic Acid-Base Equilibria/P 2 Weak acid equilibria To calculate the pH of 0.050F of o -hydroxybenzoic acid (K a =1.07 × 10 -3 ) OH COOH OH COO - + H + Initial conc. (M) 0.050 0 0 Final conc. (M) 0.050-x x x We assume that there is negligible contribution of H + from dissociation of water. ] [ ] ][ [ HA A H K Since a - + = x x - = × - 050 . 0 10 07 . 1 2 3 0 10 35 . 5 ) 10 07 . 1 ( 5 3 2 = × - × + - - x x This is a quadratic equation: ax 2 +bx+c=0 a ac b b x 2 4 2 - ± - = ) 1 ( 2 ) 10 35 . 5 )( 1 ( 4 ) 10 07 . 1 ( 10 07 . 1 5 2 3 3 - - - × - - × + × - = The negative root is rejected . 3 10 80 . 6 - × = Chem215/P.Li/Monoprotic Acid-Base Equilibria/P 3 Weak acid equilibria The amount of H + , which is similar to [OH - ], contributed from dissociation of water is very small and the above approximation is justified. ] [ ] [ + - = H K OH Since W 3 14 10 80 . 6 10 0 . 1 - - × × = M 12 7 10 4 . 1 - × = x H pH log ] log[ - = - = + 17 . 2 = If the concentration of the acid is larger (e.g. 5.0F) and/or Ka is smaller, another approximation: 5.0-x 5.0 could also be made to avoid solving the quadratic equation. Chem215/P.Li/Monoprotic Acid-Base Equilibria/P 4 Fraction of dissociation of a weak acid acid the of ion concentrat formal base conjugate d dissociate of amount = a F x HA A A = + = - - ] [ ] [ ] [ So 13.6% of the 0.050F acid has dissociated to give H + , leading to a rise in [H + ], or a decrease in pH. M

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s_monoprotic - Monoprotic acid-base equilibria Weak acid...

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