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^(o+)+Cl^(Θ), :. [H^(o+)]=[HCl]=3xx10^(-3) M, pH=-log (3xx10^(-3))=2.52`
b. `NaOH+aq rarr Na^(o+)+[overset(Θ)(O)H]`
`:. [overset(Θ)(O)H]=5xx10^(-3) M, [H^(o+)]=10^(-14)//(5xx10^(-3))`
`=2xx10^(-12) M`
`pH=-log (2xx10^(-12))=11.70`
Alternatively
`[overset(Θ)(O)H]=5xx10^(-3)`
`pOH=-log (5xx10^(-3))=2.3`
`pH=14-2.3=11.70`
c. `HBr+aq rarr H^(o+)+Br^(Θ)`,
`:. [H^(o+)]=2xx10^(-3), M, pH=-log (2xx10^(-3))=2.70`
d. `KOH+aq rarr K^(o+)+overset(Θ)(O)H, :. [overset(Θ)(O)H]=2xx10^(-3)M`,
`[H^(o+)]=10^(-14)//(2xx10^(-3)), 5xx10^(-12)`
`pH=-log(5xx10^(-12))=11.30`
Alternatively
`[overset(Θ)(O)H]=2xx10^(-3)`
`pOH=-log (2xx10^(-3))=2.7`
`pH=14-2.7=11.30`