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 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
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!!!
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