Calculate the pH of the buffer solution resulting from mixing 250 cm^3 of 0.3 moldm^-3 ethanoic acid with 250cm^3 of 0.2 moldm^-3 sodium hydroxide. The Ka of ethanoic acid is 1.8 x 10^-5.

As an acid is being added to a base, you first need to calculate what is left after they react. Without doing any calculations, we can already see that the acid is in excess as both solutions have the same volume but the acid has a higher concentration. Writing out the equation:CH₃COOH + NaOH --> Na⁺ CH₃COO^- + H₂Oshows that all reactants and products are in a 1:1 molar ratio. Firstly, you need to convert the concentrations of each reactant into moles, and consider that the reaction is limited by the amount of sodium hydroxide. It is also important to remember that the volumes must be in dm³ here, and cm³ can be converted by dividing by 1000. Moles of CH₃COOH = 0.250.3 = 0.075 molMoles of NaOH = 0.250.2 = 0.050.075 - 0.05 = 0.025Therefore 0.05 moles of the salt are formed and there are 0.025 moles of acid remaining. Now consider the equation for the dissociation of the acid:CH₃COOH <--> CH₃COO^- + H⁺and note that it is reversible, hence the need for Ka. Here the key is to notice that the conjugate pair of the acid molecule is the same ion as the one present when the salt Na⁺CH₃COO^- dissociates, and the salt will be fully dissociated in water. Recall that the equation linking these concentrations to Ka is:Ka = [H⁺][A^-]/[HA]where A^- is the anion and HA is the acid molecule. Since Ka is low, it is reasonable to assume that the vast majority of A^- ions are present because of the salt previously formed. It is also reasonable to assume that [HA] is equal to the concentration of acid molecules, but you must remember that this is not 0.3, as some of the molecules have reacted to form the salt. Since the solutions were added together, the new volume is 0.5 dm³, which leads to[HA] = 0.025/0.5 = 0.05 moldm^-3[A^-] = 0.05/0.5 = 0.1 moldm^-3We can now rearrange to find [H⁺]. Recalling that pH = -log[H⁺], we substitute in the expression found from the Ka equation:pH = -log(Ka[HA]/[A^-])Substituting in values for all variables on the right gives:pH = -log(1.8*10^-5*0.05/0.1) = 5.0457...Therefore, the pH of the solution is 5.05 to 3 s.f.