In a titration, 45.0 cm^(3) of 0.100 mol dm^(-3) sodium hydroxide solution is exactly neutralised by 40.0 cm^(3) of a dilute hydrochloric acid solution. Calculate the concentration of the hydrochloric acid solution in mol dm^(-3).

First of all, summarise everything that you know in a table or in brief notes:NaOH, 45 cm^(3), 0.1 mol dm^(-3)HCl, 40 cm^(3)The first thing that you need to do is to find the number of moles of NaOH. This is done by using the relationship of number of moles = concentration x volume.n(NaOH) = (45/1000) x 0.1 (divided by 1000 to convert from cm^3 to dm^3)n(NaOH) = 0.045 x 0.1 = 0.0045 molThe second step is to work out the reacting ratios of NaOH and HCl. This is done from the balanced equation, which in this case is NaOH(aq) + HCl(aq) --> NaCl(aq) + H₂O(aq)This shows a ratio of 1:1 which means that 0.0045 mol of NaOH will react with 0.0045 mol of HCl.The final step is to then use the relationship of number of moles = concentration x volume rearranged to calculate the concentration of the HCl solution. number of moles / volume = concentration0.0045 / (40/1000) = 0.0045 / 0.04 = 0.1125 mol dm^(-3)This figure needs to then be rounded to three significant figures, so the answer is 0.113 mol dm^(-3).