Calculate the `pH` of the following solutions:
a. `2 g` of `TlOH` dissolved in water to give `2` litre of solution.
b. `0.3 g` of `Ca(OH)_(2)` dissolved in water to give `500 mL` of solution.
c. `0.3 g` of `NaOH` dissolved in water to give `200 mL` of solution.
d. `1 mL` of `13.6 M HCl` is duluted with water to give `1` litre of solution.

Explanation

(a) `[TIOH]=(2)/(221.37xx2)=4.517xx10^(-3)`
`("mol.mass" TIOH = 2213.7)`
`:. pOH = [OH^(-)]`
`= -log 4.517xx10^(-3)=2.3452`
`:. pH=11.6548`
(b) `[Ca(OH)_(2)]=(0.3xx1000)/(74xx500)`
`=8.108xx10^(-3)M`
`:. [OH^(-)]=2xx8.108xx10^(-3)M`
`:. pOH= -log[OH^(-)]`
`= -log 16.216xx10^(-3)= 1.7900`,
`:. pH=12.21`
(c ) `[NaOH]= (0.3xx1000)/(40xx200)`
`=3.75xx10^(-2)M`
`:. [OH^(-)]=3.75xx10^(-2)M`,
`pOH= -log 3.75xx10^(-2)= 1.426`,
`:. pH=12.574`
(d) `[HCI] "formed" = (1xx13.6)/(1000)=13.6xx10^(-3)M`
`(M_(1)V_(1)=M_(2)V_(2))`
`:. [H^(+)]=13.6xx10^(-3)`,
`pH= -log 13.6xx10^(-3)`
`pH= 1.8665`