How do buffer solutions work and how do you calculate the pH of a buffer solution?

First consider what a buffer is and what a buffer can be made from. A buffer is a solution resisting change in pH formed from a weak acid and its salt (a conjugate acid base pair), or from any weak acid and a weak base. Now we know how buffers work we can see that in setting up an equation (for example for ethanoic acid), it would act as a buffer as follows:CH3COOH<> CH3COO- + H+ Adding more ethanoic acid would push the equilibrium to the right, dissociating into its constituent salt and hydrogen ion, making the solution less acidic, and adding more alkali (or salt), would react with the H+ ions to push the equilibrium to the left making the solution more acidic, therefore resisting change to pH. Because we know ethanoic acid dissociates, if it was in a buffer solution we could use Ka (the dissociation constant) to work out how much it has dissociated, therefore how many hydrogen ions are present in the solution to measure the pH. Take the example:A buffer solution was prepared which had a concentration of 0.20 mol dm–3 in ethanoic acid and 0.10 mol dm–3 in sodium ethanoate. If the Ka for ethanoic acid is 1.74 x 10–5 mol dm–3, calculate the theoretical hydrogen ion concentration and pH of the buffer solution.Ka = [H+(aq)] x [CH3COO-(aq)]/[CH3COOH(aq)]1.74 x 10–5 = [H+(aq)] x 0.10 / 0.20[H+(aq)] = 1.74 x 10–5 x 0.20/0.10 = 3.48 x 10–5 mol dm–3pH = –log(3.48 x 10–5) = 4.46

Related Chemistry A Level answers

All answers ▸

X, a gas, has a mass of 0.270g and is present in a gas syringe with a volume of 105.0cm^3 at 97C and 100kPa. Calculate the Mr of X. (5 marks)


When 80.0cm^3 of 0.500 M hydrochloric acid was added to 1.75g of impure CaCO3, not all HCl reacts. The unreacted HCl required 22.4 cm^3 of a 0.500 M solution of NaOH for complete reaction. Calculate percentage by mass of CaCO3 in the impure sample.


24.3cm^3 of 0.02moldm-3 KMnO4 reacted with 20cm^3 of iron (II) solution. Calculate the molarity of the iron (II) solution.


Explain briefly how instantaneous dipole - induced dipole bonds form.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences