A) The Ka value for acetic acid, CH3COOH(aq), is 1.8× 10–5. Calculate the pH of a 1.20 M acetic acid solution.
B)Calculate the pH of the resulting solution when 3.00 mL of the 1.20 M acetic acid is diluted to make a 250.0 mL solution.
B)Calculate the pH of the resulting solution when 3.00 mL of the 1.20 M acetic acid is diluted to make a 250.0 mL solution.
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Ka = [H+][acetate ion]/[ethanoic acid]
and since [H+] = [acetate ion] we can simplify to
Ka = [H+]2/[ethanoic acid]
then re-arrange to [H+]2 = Ka x [ethanoic acid]
so [H+]2 = 1.8 x 10^-5 x 1.2
= 2.16 x 10^-5
so [H+] = 4.65 x 10^-3
then use -log [H+] to find pH
for part B
3 x 10^-3 x 1.2 = 3.6 x 10^-3 moles in 250 ml
or 1.44 x 10^-2 moles per litre
then do the same as above, with the new concentration of ethanoic acid.
and since [H+] = [acetate ion] we can simplify to
Ka = [H+]2/[ethanoic acid]
then re-arrange to [H+]2 = Ka x [ethanoic acid]
so [H+]2 = 1.8 x 10^-5 x 1.2
= 2.16 x 10^-5
so [H+] = 4.65 x 10^-3
then use -log [H+] to find pH
for part B
3 x 10^-3 x 1.2 = 3.6 x 10^-3 moles in 250 ml
or 1.44 x 10^-2 moles per litre
then do the same as above, with the new concentration of ethanoic acid.