Calculating pH

Hi, I have been doing some chemistry past papers for the upcoming exams and have come across a question that I don't seem to be able to answer despite it only giving 2 marks.

The question is a two parter, the first part I can do it is second part I am struggling with.

The question goes like this:

A) Methanoic acid and methanoates together form buffer solutions.
(i) Calculate the pH of a solution containing equal amounts of methanoic acid and sodium methanoate.
Ka = 1.7x10^-4moldm^-3

So here I worked out pH to be -log(Ka) which gave me 3.77, the next bit is where I am stuck.

(ii) Enough sodium hydroxide is added to react with half the methanoic acid in the mixture of methanoic acid and soidum methanoate in (i).

Calculate the pH of the resulting solution.

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Does anybody have any ideas on how to answer the question and questions like this?

(edited 9 years ago)

Reply 1

lerjj

The pH of the products of a neutralization reaction is 7. The volume will have increased by a set amount.

Reply 2

Goods

ka=[h+][a-]/[ha] assume that in a buffer [a-] and [h+] are constant and in large excess.

therefore ka[ha]/[a-]=[h+] therefore pH= -log(ka[ha]/[a-])

now use that idea in part 1 [a-]=[ha] therefore the pH= -log(ka)

in part 2 half the [ha] has reacted therefore we have half the concentration of [ha] and 3[a-]/2 therefore the pH= -log(ka[ha]/3[a-]) = -log(ka/3) ([a-] and [ha] the original concentrations cancel)

Reply 3

LearnIt

Original post by Goods

Okay, so I have the correct answer now (4.25). I understand that the concentration of [HA] is halved, but I am not sure what the 3[A-]/2 is doing. (Where has the 3 come from?)

Reply 4

mynameisntbobk

Original post by LearnIt

Okay, so I have the correct answer now (4.25). I understand that the concentration of [HA] is halved, but I am not sure what the 3[A-]/2 is doing. (Where has the 3 come from?)

Think about it

Originally you had x amount of acid and its salt.
Half of the acid has reacted.. so that same amount of the salt has formed.

Originally you had x amount of the salt, now you have x + 0.5x = 1.5x = 3x/2 where x is [A-] in this case.

Reply 5

blossomx

HA : A- is 1:1
half of the acid (HA) is neutralised to becomes salt (A-)
Therefore 0.5 acid left,and 0.5 salt has formed

Acid: 1-0.5 = 0.5
Salt: 1+0.5 = 1.5

Put those ratios in and you get pH = 4.25