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

(i)0.3 g of `Ca(OH)_2` in 500 mL solution
Concentration of solution = `0.3/7.4xx1000/500` (Molar mass of `Ca(OH)_2` =74u)
=0.008 M
Since `Ca(OH)_2` is completely ionized :
`Ca(OH)_2 hArr Ca^(2+) + 2OH^(-)`
`therefore [OH^-]=2xx0.008=1.6xx10^(-2)` M
`[H_3O^+]=K_w/([OH^-]) =(1.0xx10^(-14))/(1.6xx10^(-2))=6.25xx10^(-13)` M
`pH=-log [H_3O^+]=-log (6.25xx10^(-13))`
=-log 6.25+13 =-0.796 +13 =12.204
(ii) 1 mL of 13.6 M HCl is diluted with water to give 1 L solution .
Molarity of the diluted solution :
`M_1V_1=M_2V_2`
13.6 M x 1 mL = `M_2xx1000` mL
`therefore M_2=13.6/1000=1.36xx10^(-2)` M
pH=-log `[H_3O^+]=-log (1.36xx10^(-2))`
=-log 1.36+2=-0.134+2 =1.866
(iii) 2 g TIOH dissolved in 2L solution
Molar mass of TIOH =204+16+1=221 u
Conc. of solution =`(2xx1)/(221xx2)`
`=4.52xx10^(-3)` M
Since TIOH dissociates completely as `TIOH hArr Tl^(+) + OH^(-)`
`therefore [OH^-]=4.52xx10^(-3)` M
`[H_3O^+]=K_w/([OH^-])=(1.0xx10^(-14))/(4.52xx10^(-3))=2.212xx10^(-12)` M
pH=-log `[H_3O^+]=-log (2.212 xx10^(-12))`
=-(0.345-12)=11.655
(iv) 0.3 g of NaOH dissolved in 200 mL of solution
Conc. of solution `=0.3/40xx1000/200`
=0.0375 M (`because` Molar mass of NaOH = 40)
Since NaoH dissociates completely as :
`NaOH hArr Na^(+)+OH^(-)`
`[OH^-]`=0.0375 M
`[H_3O^+]=K_w/([OH^-])=(1xx10^(-14))/0.0375=2.67xx10^(-13)`
`pH=-log [H_3O^+]=-log (2.67xx10^(-13))`
=-(4.265-13)=12.57