In an experiment to find the concentration of calcium hydroxide in limewater , 25.0 cm3 of limewater needed 18.8 cm3 of 0.0400 mol/dm3 hydrochloric acid to neutralise it.

0.01504 mol/dm3

Explanation:

Calcium Hydroxide - Ca(OH)2

Hydrochloric acid - HCl

The reaction between both is given as;

2HCl   +   Ca(OH)2   --> CaCl2  + 2H2O

2   :  1

The relationship between the concentration and volume is given as;

CaVa  /  CbVb = na  /  nb

where a = acid (HCl) and b = base Ca(OH)2

Inserting the values;

0.04 * 18.8  /  (Cb * 25)  =  2 / 1

0.752 / 25Cb = 2

Cb = 0.752 / 50  = 0.01504