Calculate the pH of 0.003 M HCl.

Explanation

Since HCl ionises completely. Therefore,
`[H_(3)O^(+)] = [HCl] = 0.003 M = 3 xx 10^(-3)M`
`pH = -log[H_(3)O^(+)] = -log[3 xx 10^(-3)] = -log 3 + 3 log 10 [because log 10 = 1, log 3 = 0.4771]`
`= -0.4771 + 3 = 2.5229`