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.

Answer

a) `11.65` , b) `12.21`, c) `12.57` , d) `1.87`

Explanation

(a) For 2g of TlOH dissolved in water to give 2 L of solution:
`[TIOH_((aq))] = (2)/(2)g//L`
`= 2/2 xx 1/(221) M`
`= 1/(221) M`
`TlOH_((aq)) rarr Tl_((aq))^(+) + OH_((aq))^(-)`
`[OH_((aq))^(-)] = [TlOH_((aq))] = 1/(221) M`
`K_(w) = [H^(+)] [OH^(-)]`
`10^(-4) = [H^(+)] ((1)/(221))`
`221 xx 10^(-14) = [H^(+)]`
`rArr pH = - "log" [H^(+)] = - "log" (221 xx 10^(-14))`
`= - "log" (221 xx 10^(-12))`
`= 11.65`
(b) For `0.3 g` of `Ca(OH)_(2)` dissolved in water to give mL of solution:
`Ca(OH)_(2) rarr Ca^(2) + 2OH^(-)`
`[Ca(OH)_(2)] = 0.3 x (1000)/(500) = 0.6 M`
`[OH_((aq))^(-)] = 2 xx [Ca(OH)_(2aq)] = 2 xx 0.6`
`= 1.2 M`
`[H^(+)] = (K_(w))/([OH_(aq)^(-)])`
`= (10 - 14) M`
`= 0.833 xx 10^(-14)`
`pH = - "log" (0.833 xx 10^(-14))`
`= - "log" (8.33 xx 10^(-13))`
`= (-0.902 + 13)`
`= 12.098`
(c) For `0.3 g` of `NaOH` dissolved in water to give 200 mL of solution.
`NaOH rarr Na_((aq))^(+) + OH_((aq))^(-)`
`[NaOH] = 0.3 xx (1000)/(200) = 1.5 M`
`[OH_(aq)^(-)] = 1.5 M`
Then, `[H^(+)] = (10^(-14))/(1.5)`
`= 6.66 xx 10^(-13)`
`pH =- "log" (6.66 xx 10^(-13))`
`= 12.18`
(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:
`13.6 ×x 1 mL = M_(2) × 1000 mL`
(Before dilution) (After dilution)
`13.6 ×x 10^(-3) = M_(2) × 1L`
`[H^(+)] = 1.36 × 10^(-2)`
`pH = - "log" (1.36 × 10^(-2))`
`= (– 0.1335 + 2)`
`= 1.866 ∼ 1.87`