Sodium Hydroxide can react with sulfuric acid in an acid base reaction. Outline the equation for this reaction, name the product. Finally a titration reaction is conducted. Determine the mass of NaOH needed to neutralise 12.4cm^3 of 0.10moldm^-3 H2SO4.

The equation is outlined below:      2NaOH + H₂SO₄------->> Na₂SO₄ + 2H₂OAs explained in the question this is an acid base reaction, where the sodium hydroxide is the base and the sulphuric acid is the acid. This reaction produces the salt sodium sulphate. For the reaction to be balanced the sodium hydroxide and sulphuric acid is mixed in a 2:1 ratio.In order to determine the mass of the sodium hydroxide we must first of all determine the number of moles of the sulphuric acid by dividing the volume (12.4) by 1000 and multiplying by the concentration (0.10). We now have the number of moles of sulphuric acid which is 0.00124moldm^-3, using the equation used produced further on in the question we can determine the number of moles of Sodium Hydroxide by multiplying the number of moles of sulphuric acid by the stoichiometric co-efficient of the sodium hydroxide, which gives us a value of 0.00248moldm^-3 when we multiply the number of moles of sulphuric acid by 2. We can then determine the relative molecular mass of sodium hydroxide from the periodic table which is 40g/mol. To determine the mass we multiply the Mr by the number of moles to get a mass of 0.0992g.