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11 years ago
there is 200mL of 1.00M solution of acetic acid (CH3COOH) with 200mL of potassium hydroxide (KOH) of unknown concentration. The pH of the resulting mixture was 10 units higher than that of the original acid solution. Ka of CH3COOH is 1.8x10^-5. Find the unknown concentration of KOH.
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wrote...
11 years ago
when the Ka = 1.8 e-5
the pKa = 4.745

find final cnocentration
The pH of the resulting mixture was 10 units higher :  14.745
the pOH would be -0.745
[OH-] would be 10^-0.745 = 0.18 Molar

find final moles of OH
that 0.18 Molar is for a total mix of 400 ml
0.400 Litres @ 0.18 mol / litre = 0.072 moles of OH

find the moles of OH destroyed by the acid
0.200 litres of acid at 1.00 M = 0.200 moles of acid
by the equation
1 HC3H3O2  &  1 KOH Rightwards Arrow  1 K C2H3O2  &  1 H2O
0.200 moles of acid destroyed 0.200 moles of KOH

so
 0.200 moles of KOH + 0.072 moles of OH remailing = 0.272 moles OH- total
the original 0.272 moles of KOH was in 200 ml
so
0.272 moles / 0.200 L = 1.36 Molar KOH

that's likely the answer they are looking for...
however the Ka has only 2 sig figs
so your answer should really be rounded off to
1.4 Molar KOH
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