Calculate pH for:
(a) `0.001 NaOH`, (b) `0.01N Ca(OH)_(2)`,
(c ) `0.01M Ca(OH)_(2)`, (d) `10^(-8)M NaOH`,
(e ) `10^(2)MNaOH`, (f) `0.0008MMg(OH)_(2)`
Assume complete ionisation of each.

Explanation

(a) `0.001N NaOH:`
`{:(NaOHrarr,Na^(+)+,OH^(-)),(10^(-3)N,0,0),(0,10^(-3),10^(-3)):}`
`:. [OH^(-)]=10^(-3)M`
`:. pOH= -log[OH^(-)]`
`= -log 10^(-3)=3`
`:. pH=14-pOH=14-3=11`
`:. pH =11`
(b) `0.01N Ca(OH)_(2)`
`{:(Ca(OH)_(2)rarr,Ca^(2+)+,2OH^(-)),(10^(-2)N,0,0),(0,10^(-2),10^(-2)):}`
`:. [OH^(-)]=10^(-2)M`
`:. pOH=2`
`:. pH=12`
(c ) `0.01M Ca(OH)_(2)`:
`{:(Ca(OH)_(2)rarr,Ca^(2+)+,2OH^(-)),(10^(-2)M,0,0),(0,10^(-2),2xx10^(-2)):}`
`[OH^(-)]=2xx10^(-2)M`
`:. pOH=1.6989`
`:. pH=14-1.6989=12.3010`
(d) `10^(-8)MNaOH:`
`{:(NaOHrarr,Na^(+)+,OH^(-)),(10^(-8)M,0,0),(0,10^(-8),10^(-8)):}`
`:. [OH^(-)]=10^(-8)M`
Now proceed for `OH^(-)` as in problem 19 part (e ).
(e ) `10^(2)MNaOH:`
`{:(NaOHrarr,Na^(+)+,OH^(-)),(10^(2),0,0),(0,10^(2),10^(2)):}`
`[OH^(-)]=10^(2)M`
Now proceed in problem 19 part (f).
(f) `0.0008MMg(OH)_(2)`:
`{:(Mg(OH)_(2)rarr,Mg^(2+)+,2OH^(-)),(8xx10^(-4)M,0,0),(0,8xx10^(-4),2xx8xx10^(-4)):}`
`:. [OH^(-)]=16xx10^(-4)M`
`:. pOH=2.7958`
`pH=11.2041`