A solution of acetic acid and sodium acetate was prepared, by dissolving 4.1 g of sodium acetate in 750 cm^3 of 0.085 mol/dm^3 acetic acid, at 25 degrees. 10 Cm^3 of 2 mol/dm^3 HCl was added. Ka is 1.76*10^-5, calculate and explain the change in pH

The first thing to do in any calculations question is to list out what you have, and what the question is asking you. In this case, the question wants pH, which is -log[H+]. So what we need is [H+]. We have the volume and concentration of acetic acid, and the mass in grams of sodium acetate. These can both be used to calculate moles via (vol/1000) * conc, and mass in grams/molecular weight respectively. This gives0.05 moles of sodium acetate0.06375 moles of acetic acidThen, as we know Ka, or the dissociation constant, we can determine [H+] from this, using the equation, Ka= [H+][CH3COO-]/[CH3COOH]. We can use the moles of sodium acetate for the anion, as there is 1 mole of anion per mole of sodium acetate.Although the equation is in square brackets, indicating concentration, we can use moles because the division cancels out the units. However, if you feel more comfortable, concentration would be fine too.Ka= [H+][CH3COO-]/[CH3COOH][H+]= Ka*[CH3COOH]/[CH3COO-][H+]= 1.7610^-5 * 0.06375/0.05 = 2.24410^-5-log(Ans)= pH= 4.65 beforeThen adding 0.02 moles of HCl (calculated with (vol/1000)conc)) is the same as adding 0.02 moles of H+, as there is 1 mole of H+ per mole of HCl. This then reacts with the anion, also in a 1:1 reaction to decrease its moles by 0.02, and creates 0.02 moles more of the undissociated acetic acid per la chatelier’s principle. This means that pH remains relatively unchanged, as the [H+] difference is not significant enough, which is what can be expected, as the solution is a buffer, as it is made from a weak acid, and its conjugate base.Acetic acid= 0.08375 molesEthanoate anion= 0.03 moles[H+] = 1.7610^-5 * (0.08375/0.03)= 4.91*10^-5-log(ans) = pH = 4.31pH change = 0.34.