20cm3 of 0.5moldm-3 of HCL is diluted by adding 15cm3 of water. This diluted solution is titrated against a 0.3moldm-3 solution of NaOH. What is the volume of the NaOH in cm3 required to reach the endpoint of the titration?

We need to calculate how many moles of HCl are being titrated. This is the volume in dm3 times the concentration (trick: you can do dimensional analysis if unsure of the formula): number of moles of HCl = 0.02 x 0.5 = 0.01 moles of HCl. By checking the chemical equation, we see that we need the same number of moles of NaOH for titration. The endpoint is reached when all HCl is consumed, so we need to see what volume of 0.3 moldm-3 NaOH solution contains 0.01 moles of NaOH. Again, this is the same formula used above but rearranged and the result is 0.01/0.3=0.033 L which is approx. 33 cm3.Equation of chemical reaction: HCl + NaOH = NaCl + H2O