Propanoic acid is CH3CH2COOH. Let's simplify that by calling it HPr.
HPr + NaOH ==> NaPr + H2O
I have no idea what an IRE-C table is.
moles NaOH = L x M = 0.01350 x 0.200 M = 0.027
moles HPr = L x M = 0.05000 x 0.100 = 0.005.
initial concns:
HPr = 0.005 moles/0.0635 = ??
NaOH = 0.0135/0.0635 = ??
pH = pKa + log [(base)/(acid)]
pH = substitute what you are using for pKa (that may be the problem) and values for concn base (NaPr) and acid (HPr). If this doesn't work out for you, repost but give the value you are using for Ka and pKa.
A buffer solution is made as followed:
i)adding 13.50mL of 0.200mol/L sodium hydroxide to 50.00mL of 0.100mol/L propanoic acid.
ii)diluting the resulted buffer into a total volume of 100.00mL
Using IRE-C tables (if possible) calculate:
a)The pH of the buffer -before the dilution
b)The pH of the buffer- after the dilution
c) 1.50mL of 0.10mol/L hydrochloric acid solution is added to 25.00mL of the diluted buffer solution. Calculate the pH after the addition of the hydrochloric acid solution.
I did part a) but I am not getting the correct answer, which is given. I know that part b) is done the same way and when diluting the buffer into a total volume of 100.00mL it should not affect the pH. However, since I cannot get the right answer for part a) my answer for part b) is also incorrect.
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