Question #9075

Glycolic acid which is a monoprotic acid and a constituent in sugar has a pka of 3.9. A 25 ml solution of glycolic acid is tritrated to the stoichiometric point with 35.8 ml of 0.020 M sodium hydroxide solution. What is the ph of the resulting solution at the stoichiometric point?

Expert's answer

pK_{a} = -logK_{a}

K_{a} = 10^-3.9 = 1.26x10^-4

K_{a} = [H+][An-]/[HAn] = [H+]^2 /[HAn] "An" is HOCH_{2}COO-

NaOH + HAn = NaAn + H_{2}O

n of NaOH is:

0.020 mol is in 1000ml

x mol is in 35.8

x = 0.000716 mol

so n of HAn is the same,and concentration is 0.000716 mol in 25ml,or 0.029 mol in 1000 ml or

0.029M

Now, acetate ions are a base and will react with water by the reaction:

An^{-} + H_{2}O --> HAn + OH^{-}

K_{b} for this reaction can be calculated from K_{a} x K_{b} = K_{w} to be 7.94x10^-11

By that equation, [HAn] = [OH^{-}], so K_{b} = x^2/0.029

You can solve this to show that [OH^{-}] = 1,52x10^-6 M. pOH, then = 5.82 and pH = 14 - pOH = 8.18

K

K

NaOH + HAn = NaAn + H

n of NaOH is:

0.020 mol is in 1000ml

x mol is in 35.8

x = 0.000716 mol

so n of HAn is the same,and concentration is 0.000716 mol in 25ml,or 0.029 mol in 1000 ml or

0.029M

Now, acetate ions are a base and will react with water by the reaction:

An

K

By that equation, [HAn] = [OH

You can solve this to show that [OH

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