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Chemistry pH question

can anyone tell me how to answer this question for chem a level?

A buffer solution was made by mixing 10cm3 of sodium hydroxide with a concentration of 0.23moldm3 and 30cm3 of propanoic acid with a concentration 0.55moldm3. for propanoic acid, Ka=1.35x10^-5
What’s the question?
Original post by connief2004
can anyone tell me how to answer this question for chem a level?

A buffer solution was made by mixing 10cm3 of sodium hydroxide with a concentration of 0.23moldm3 and 30cm3 of propanoic acid with a concentration 0.55moldm3. for propanoic acid, Ka=1.35x10^-5

So you want to know the pH?

Use the balanced equation to find the moles of salt formed and the mol of propanoic acid that you end up with (it's a 1:1 reaction).

ka = [H+][A-]/[HA]

this is normally for concentrations BUT in buffers, as the components are both in the same solution you can substitute moles

This will give you the [H+] and from that you get the pH
Reply 3
Original post by Just_Bored
What’s the question?

oops lol, its whats the pH of the buffer
Reply 4
Original post by connief2004
can anyone tell me how to answer this question for chem a level?

A buffer solution was made by mixing 10cm3 of sodium hydroxide with a concentration of 0.23moldm3 and 30cm3 of propanoic acid with a concentration 0.55moldm3. for propanoic acid, Ka=1.35x10^-5

The reaction
NaOH+ CH3CH2COOH--> CH3CH2COONa + H2O

1. Convert everything to Moles
Moles of NaOH= 0.23*10/1000= 2.3*10^-3
Moles of Propanoic acid= 0.55*30/1000= 0.0165
So NaOH is limiting and propanoic acid is in excess
Therefore moles of sodium propanoate (salt) = 2.3*10^-3
Moles of propanoic acid u end up with= 0.0142 (after the salt is formed)

2. Use Ka to find [H+]
[H+]= Ka* Acid/Salt
= (1.35x10^-5)* (0.0142/2.3*10^-3)
[H+]=8.33*10^-5

3. We find the pH= -log[H+]
= -log[8.33*10^-5]
= 4.08

I hope this helps, good luck
(edited 2 years ago)

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