Titrations Involving Acids and Alkalis

Having volume/ concentration units additionally complicates such calculations.  One could convert these units to mole units and then proceed as with the other calculations.  However, there is a much easier method, which employs the following formula:

M1          x              V1           =              mol. of 1 in the balanced equation

M2          x              V2           =              mol. of 2 in the balanced equation

Were      M= molarity or mol dm^-3

                V = volume

1 and 2 refer to the acid and the alkali or vice versa.

In such questions one will have a neutralisation equation from which the values on the right hand side of the above formula are found.  The question will then provide three of the other four values. E.g. V₁, M₂ and V₂.  What you then do is to rearrange the equation in terms of the unknown - M₁ in this case.  Insert the values given and get the answer.  Now try the following questions:

(1)                 In the reaction

Ca(OH)₂ (aq)         +              2HCl (aq)               =              CaCl₂ (aq)              +              2H₂0 (aq)

It required 20cm³ of 0.5mol dm^-3 hydrochloric acid to neutralise 10cm³ of a calcium hydroxide solution.  Calculate the concentration of the alkali.  [ Hint call the alkali 1 and the acid 2 so that you require M₁]

(2)                 In the reaction

2NaOH   +              H₂SO₄     =              Na₂SO₄                   +              H₂0

2.5cm³ 0.5mol mol dm^-3 sodium hydroxide, was neutralised using 0.75 mol dm^-3 sulphuric acid.  Calculate the volume of acid required for neutralisation [ Hint: call the acid 1 and the alkali 2 so that you require V₁]

        Garreth Mc Bride: Chemistry Department: Loreto College, Coleraine.