Assuming complete dissociation, calculate the `pH` of the following solutions,
a. `0.003 M HCl, b. 0.005 M NaOH`,
c. `0.002 M HBr, d. 0.002 M KOH`

Explanation

(a) `HCl + aq rarr H^(+) + Cl^(-) , :. [H^(+)]=[HCl]=3xx10^(-3)M, pH = - log (3xx10^(-3))=2.52`
(b) `NaOH + aq rarr Na^(+) + OH^(-)= 5xx10^(-3) M, [H^(+)]=10^(-14)//(5xx10^(-3))=2xx10^(-12)M`
`pH = - log (2xx10^(-12))=11.70`
(c) `HBr + aq rarr H^(+) + Br^(-), :. [H^(+)]=2xx10^(-3)M, pH = - log (2xx 10^(-3))=2.70`
(d) `KOH + aq rarr K^(+) + OH^(-) , :. [OH^(-)]=2xx10^(-3)M,[H^(+)]=10^(-14)//(2xx10^(-3))=5xx10^(-12)`
`pH = - log (5xx10^(-12))=11.30`