# How do I answer this hard acids and bases titration questions?

I had it in my mock paper today.
It was something along the lines of
butanoic acid Ka= 1.51 x 10-5
25cm^3 0.1 moldm-3 butanoic acid 20cm3 reacted with 0.1moldm-3 NaOH calculate the pH of the solution

I got an answer of 5 ish
I got 5.42 idk where I went wrong
Original post by Meepmop123
I had it in my mock paper today.
It was something along the lines of
butanoic acid Ka= 1.51 x 10-5
25cm^3 0.1 moldm-3 butanoic acid 20cm3 reacted with 0.1moldm-3 NaOH calculate the pH of the solution

I got an answer of 5 ish
Step 1 is to find the concentration of the products.
The reaction is 1:1
This means that the butanoic acid is in excess and all of the sodium hydroxide is used up.
This gives a buffer mixture of butanoic acid (excess) and sodium butanoate.
Apply the acid dissociation equation:
ka = [H+][A-]/[HA]

where [A-] is the concentration of the salt formed and [HA] is the concentration of the unreacted acid.
Solve for [H+]

mol A- = 0.02 x 0.1 = 0.002
mol HA = 0.005 x 0.1 = 0.0005

1.51 x 10-5 = [H+] x 0.002/0.0005

[H+] = 1.51 x 10-5 /4 = 3.775 x 10-6

pH = 5.42

You are correct!
Original post by charco
Step 1 is to find the concentration of the products.
The reaction is 1:1
This means that the butanoic acid is in excess and all of the sodium hydroxide is used up.
This gives a buffer mixture of butanoic acid (excess) and sodium butanoate.
Apply the acid dissociation equation:
ka = [H+][A-]/[HA]

where [A-] is the concentration of the salt formed and [HA] is the concentration of the unreacted acid.
Solve for [H+]

mol A- = 0.02 x 0.1 = 0.002
mol HA = 0.005 x 0.1 = 0.0005

1.51 x 10-5 = [H+] x 0.002/0.0005

[H+] = 1.51 x 10-5 /4 = 3.775 x 10-6

pH = 5.42

You are correct!
Oh my GOD that’s amazing !!
I had a mock exam today and all the clever students got pH of 4 and I was so nervous about my answer being incredibly wrong!!!
Original post by charco
Step 1 is to find the concentration of the products.
The reaction is 1:1
This means that the butanoic acid is in excess and all of the sodium hydroxide is used up.
This gives a buffer mixture of butanoic acid (excess) and sodium butanoate.
Apply the acid dissociation equation:
ka = [H+][A-]/[HA]

where [A-] is the concentration of the salt formed and [HA] is the concentration of the unreacted acid.
Solve for [H+]

mol A- = 0.02 x 0.1 = 0.002
mol HA = 0.005 x 0.1 = 0.0005

1.51 x 10-5 = [H+] x 0.002/0.0005

[H+] = 1.51 x 10-5 /4 = 3.775 x 10-6

pH = 5.42

You are correct!
Tbh we haven’t even learnt buffers yet.
I just used one of those equilibrium tables to work out the moles and then I used Ka