10cm^3 of 1M NaOH solution is mixed with 15cm^3 of 0.5M HCl, what is the resulting pH of the solution?

Step 1: Work out the moles of NaOH and HCl added into the solution 10cm^3 of 1M NaOH would mean that there is 10cm^3/1000cm^3 x 1M present, i.e. 0.01 moles 15cm^3 of 0.5M HCl would mean that there is 15cm^3/1000c^3 x 0.5M present, i.e. 0.0075 moles Step 2: Work out the moles after they have reacted. HCl and NaOH react 1:1 and you can assume that in these conditions they do so to completion (self ionization of water = 10^-14) meaning that there is 0.01 - 0.0075 = 0.0025 moles. NaCl is soluble and does not affect the pH. Step3: Work out the pH of a solution of 0.0025 moles of NaOH in 25cm^3. 0.0025moles / 25cm^3 x 1000cm^3 = 0.1M (moles/dm^3). To calculate the pH there are two ways about it. The first and possibly simpler is to use the relation that pOH + pH = 14 and pOH = - log([OH-]). In this case pOH = - log(0.1) = 1 and therefore the pH = 14 - 1 = 13. The second method is to work out the concentration of H+ (H3O+) using [OH-][H+] = 10^-14 = Kw (self ionisation of water constant) and pH = -log([H+]). [H+] = 10^-14 / 0.1 = 10^-13M, then pH = -log( 10^-13 ) = 13.