Q1. Two beakers, A and B, each contain 100.0 cm^3 of 0.0125 mol/dm^3 nitric acid. Calculate the pH of the solution formed after 50.0 cm^3 of distilled water are added to beaker A. Give your answer to 2 decimal places.

A1.   First it is important to know that nitric acid (HNO3) is a strong acid and so [HNO3] = [H+]    Beaker A contains 100cm^3 = 100ml   Final volume is 150ml (50cm^3 of distilled water is added to beaker A)   So, (100ml/150ml) x 0.0125 mol/dm^3 will give a [H+] as 8.3 x 10^-3   as pH = -log[H+], pH = 2.08