A pH 3.0 solution of HCl(aq) is diluted by adding water to produce a pH 5.0 solution - HSC - SSCE Chemistry - Question 15 - 2013 - Paper 1

To calculate the volume of water needed to achieve a pH of 5.0 starting from a pH of 3.0, we can use the relationship between pH and hydrogen ion concentration:

The formula for pH is given by: $ext{pH} = - ext{log}[H^+]$

For pH 3.0, the hydrogen ion concentration is: $[H^+] = 10^{-3} ext{ mol/L}$

And for pH 5.0, it is: $[H^+] = 10^{-5} ext{ mol/L}$

To dilute HCl from pH 3.0 to pH 5.0, the dilution factor required is: $ext{Dilution Factor} = \frac{[H^+](3.0)}{[H^+](5.0)} = \frac{10^{-3}}{10^{-5}} = 100\text{ times}$

Assuming we take 10 mL of the pH 3.0 solution, we need 990 mL of water to dilute it, leading us to the following total volumes:

$ext{Total Volume} = ext{Original Volume} + ext{Water Volume} = 10 ext{ mL} + 990 ext{ mL} = 1000 ext{ mL}$

Thus, the correct choice is (C): 10 mL of the original solution and 990 mL of water added.