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
TD
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