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) pH of 2 g of TlOH in 2 litre of solution
Molarity of TlOH solution `=("Mass of TiOH" // "Molar mass")/("Volume of solution in titres") =((2g)//(204+17 g" mol"^(-1)))/(2L)`
`=4.525 xx 10^(-3) " mol "L^(-1) =4.525 xx 10^(-3)M`
`TIOH(s) overset((aq))(to) TI^(+) (aq) +OH^(-) (aq)`
`4.525 xx10^(-3)M`
`[H_(3)O^(+)] =(K_(w))/[[OH^(-)]] =((10^(-14)M^(2)))/((4.525 xx 10^(-3)M)) =2.21 xx 10^(-12)M`
`pH =- log [H_(3)O^(+)] =- log(2.21 xx 10^(-12))`
`=- (log 2.21 log 10) =(12- log 2.21) =12-0.344 =11.656`
(b) pH of 0.3 g of `Ca(OH)_(2)` in 500 mL of solution
Molarity of `Ca(OH)_(2) = ("Mass of" Ca(OH)_(2)//"Molar mass")/("Volume of solution in litre") =((0.3 g)//(74 g" mol"^(-1)))/(0.5L)`
`=8.1 xx 10^(-3) " mol " L^(-1) =4.525 xx 10^(-3)M`
`Ca(OH)_(2) overset((aq))(to) Ca^(2+) (aq) +2OH^(-) (aq)`
`8.1 xx 10^(-3)M 16.2 xx 10^(-3)M`
`pOH =- log [OH^(-)] =- log (16.2 xx 10^(-3))`
`=- (log 16.2 - 3 log 10) =(3- log 16.2)`
`=(3-1.209) = 1.791`
`pH =14 -pOH =14- 1.791 = 12.21`
(c) pH of 0.3 g of NaOH in 200 mL of solution
Molarity of NaOH solution `=("Mass of NaOH"//"Malar mass")/("Volume of solution in litres") =((0.3g)//(40 g" mol"^(-1)))/(0.2 L)`
`=0.0375 "mo" L^(-1) =0.0375 M`
`NaOH overset((aq))(to) Na^(+) (aq) +OH^(-) (aq)`
`0.0375 M 0.0375 M`
`pOH =- log [OH^(-)] =- log (3.75 xx 10^(-2))`
`=- (log 3.75 -2 log 10) =(2-log 3.75) =(2 -0.574) =1.426`
`pH =14 -pOH =14 -1.426 =12.574`
(d) pH of 1 mL of 13.6 M HCI solution diluted to 1L
Molarity of diluted may be calculated as :
`(13.6 M) xx (1mL) =M_(2) xx (1000 mL ) " or " M_(2) =((13.6 M) xx (1 mL))/((1000 mL))`
`=1.36 xx 10^(-2) M`
`HCI to H^(+) (aq) + CI^(-) (aq)`
`1.36 xx 10^(-2) M 1.36 xx 10^(-2)M`
`pH =- log [H^(+)] =- log (1.36 xx 10^(-2))`
`=- (log 1.36 -2 log 10) =(2- log 1.36)`
`=(2 - 0.1335) = 1.867`