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) Molar conc. Of TlOH `= (2g)/((204+16+1)g "mol"^(-1))xx(1)/(2L)=4.52xx10^(-3)M`
`:. [OH^(-)]=[TlOH]=4.52xx10^(-3)M`
`[H^(+)]=10^(-14)//(4.52xx10^(-3))=2.21xx10^(-12)M`
`pH = - log (2.21xx10^(-12))=12-(0.3424)=11.66`
(b) Molar conc. of `Ca(OH)_(2) = (0.3 g)/((40+34) g "mol"^(-1))xx(1)/(0.5L)=8.11xx10^(-3)M`
`Ca(OH)_(2) rarr Ca^(2+)+ 2 OH^(-)`
`:. [OH^(-)]=2[Ca(OH)_(2)]=2xx(8.11xx10^(-3))M = 16.22xx10^(-3)M`
`pOH = - log (16.22xx10^(-3))=3-1.2101=1.79`
`pH = 14-1.79=12.21`
(c) Molar conc. of NaOH `=(0.3 g)/(40 g "mol"^(-1))xx(0.2L)=3.75xx10^(-2)M`
`[OH^(-)]=3.75xx10^(-2)M`
`pOH = - log (3.75xx10^(-2))=2-0.0574 = 1.43 :. pH = 14 - 1.43 = 12.57`
(d) `M_(1)V_(1) = M_(2)V_(2) :. 13.6 M xx 1 mL = M_(2) xx 1000mL :. M_(2) = 1.36xx10^(-2)M`
`[H^(+)]=[HCl]=1.36xx10^(-2)M, pH = - log (1.36xx10^(-2))=2-0.1335 ~= 1.87`