The Ka of 1.75x10^-5. Calculate the pH when 50mL of 0.10 M acetic acid and 50mL 0.10M sodium acetate are mixed. (I got 4.76 but not sure if this is right). Calculate pH when 0.0025 mol of NaOH is added to the previous solution, assuming the volume doesn't change. Comment on the effectiveness of the solution as a buffer.

The acetic acid provides the CH3COOH in the equilibrium, and sodium acetate provides the CH3COO- in the equilibrium:

CH3COOH(aq) + H2O(l) <--> CH3COO-(aq) + H3O+(aq)

We know [CH3COOH] = 0.10 M (but there is a dilution when the two things are combined, because you are moving from 50 mL to 100 mL total volume)

So this means [CH3COOH] in end solution = 0.05 M (because we have diluted by a factor of two)

Similarly, the acetate is diluted so [CH3COO-] in end solution = 0.05 M

Ka = [CH3COO-][H3O+] / [CH3COOH]

1.75 x 10^-5 = (0.05) [H3O+] / (0.05)

This means [H3O+] = 1.75 x 10^-5 M

pH = -log[H3O+]

pH = -log(1.75 x 10^-5)

pH = 4.76

Now next part:

I'm not 100% sure if you are allowed to use this in the syllabus but I can't think of any other way to do it, but there is something called the Hendersen-Hasslebalch equation which is used for predicting pH in buffers after stuff is added

The equation is: pH = pKa + log10([A^-]/[HA])

A^- here is CH3COO-

HA here is CH3COOH

pKa is just -log(Ka) = -log(1.75 x 10^-5) = 4.756....

So 0.0025 mol of NaOH has conc of 0.0025 / 0.100 = 0.025 M

This NaOH is going to react with the acetic acid: CH3COOH + NaOH --> NaCH3COO + H2O

So this means CH3COOH will reduce in conc by 0.025 M. So it goes from 0.05 M to 0.05 - 0.025 = 0.025 M

Now at the same time you made NaCH3COO so this one has to go up. Therefore 0.05 M goes to 0.05 + 0.025 = 0.075 M

Now sub in our equation:

pH = pKa + log10([A^-]/[HA])

pH = 4.756... + log(0.075/0.025)

pH = 5.23

Therefore, the solution is effective in minimising pH change as it only changed by 0.5 units approximately