Calculate the pH for
a. `0.001 M NaOH`
b. `0.01 M Ca(OH)_(2)`
c. `0.0008M Ba(OH)_(2)`
d. `0.004 M NaOH`

Explanation

`pOH=-log[OH^(-)]`
`pH=14-pOH`
a. `pOH=-log[OH^(-)]=-log[1xx10^(-3)]=3`
`pH=14-3=11`
b. `0.01M Ca(OH)_(2)`
Conc. Of `[OH^(-)]=0.02`
`:.` 1 mol `Ca(OH)_(2)` gives 2 mol `OH^(-)`
`poH=-log[2xx10^(-2)=1.6990`
`pH=14-1.6990=12.3010`
c. `underset(0.0008)(Ba(OH)_(2))toBa^(2+)+underset(0.0016)(2OH^(-))`
`pOH=-log[1.6xx10^(-3)]=2.7959`
`pH=14-2.7959=11.2041`
d. `underset(0.004)(NaOH)toNa^(+)+underset(0.004)(OH^(-))`
`pOH=-log(4xx10^(-3))=2.3980`
`pH=14-2.3980=11.6020`