# chemistry

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24 mL of 0.39 mol/L acetic acid is titrated with a standardized 0.33 mol/L KOH solution. Calculate the pH of the solution after 17 mL of the KOH solution has been added. Assume the Ka of acetic acid is 1.8 x 10-5.

• chemistry -

This is a buffered solution. Calculate mols acetic acid and mols KOH; you will have some salt produced as well as some acetic acid remaining. Then pH = pKa + log[(base)/(acid)]. Post your work if you get stuck.

• chemistry -

mols acetic acid = 0.024 Lx 0.39 mol/L
= 0.00936 mols

mols KOH = 0.33 mol/L x 0.017 L
= 0.00561 mols

pH = +log(0.00936)(0.00561)
= -4.279
= 14 + -4.279
=9.7

I am stuck

• chemistry -

KOH + CH3COOH ==> CH3COOK + HOH

mols KOH = 0.00561. correct.
mols CH3COOH = 0.00936. correct.
But from here on is new territory.
You must recognize that KOH and CH3COOH are strong base and weak acid (which you probably know already) so what happens is these react to produce how much of the salt? It will produce, of course, the lesser of the two, which in this case is 0.00561 mols of the CH3COOK. So how much of the CH3COOH is used? Obviously 0.00561 mol has been used. How much of the CH3COOH is left unreacted? The difference which is 0.00936 - 0.00561 = 0.00375.
So you see there is CH3COOH unreacted PLUS the salt of the weak acid which makes this a buffered solution. That is solved by the Henderson-Hasselbalch equation.
Concn CH3COOH = mols/L (L here will be 17 mL + 24 mL = 41 mL or 0.041 L).
concn CH3COOL (the salt) is mols/L.

HH equation is
pH = pKa + log [(base)/(acid)/]
The base in this case is the conjugate base of acetic acid which is the salt.