a. What amount of `H_(2)SO_(4)` must be dissolved in `500mL` of solution to have a `pH` of `2.15`?
b. What amount of `KOH` must be dissolved in `200mL` of solution to have a `pH` of `12.3`?
c. What amount of `ca(OH)_(2)` must be dissolved in `100mL` of solution to have a `pH` of `13.85`?

Explanation

a. `pH = 2.15`
`-log [H^(o+)] = 2.15`
`log [H^(o+)] =- 2.15 =- 2 - 0.15 +1 - 1 = bar(3).85`
`[H^(o+)] = Antilog (bar(3).85) = 7 xx 10^(-3) N`
strength of `H_(2)SO_(4) (gL^(-1)) = N xx Ew`
(Ew of `H_(2)So_(4) = (98)/(2) = 49`)
`= 7 xx 10^(-3) xx 49 = 0.343 g L^(-1)`
`= (0.343 xx 500 mL)/(1000mL) = 0.1715g //500 mL`
Alternatively:
Strength of `H_(2)SO_(4) (g L^(-1))`
`=M xx Mw, (M = (7 xx 10^(-3)N)/(2))`
`= (7 xx 10^(-3))/(2) xx 98 = 7 xx 10^(-3) xx 49 gL^(-1)`
`= (7 xx 10^(-3))/(2) g//500mL = 0.1715g//500mL`
b. Since `pH gt gt`( basic solution), first calculate `pOH` and `[overset(Θ)OH]`. Then calculate amount of `KOH//200mL`
`pH = 12.3, pOH = 14 - 12.3 = 1.7`
`- log [overset(Θ)OH] = 1.7`
`log[overset(Θ)OH] =- 1.7 =- 1 - 0.7 +1 - 1 = bar(2).3`
`[overset(Θ)OH] = Antilog (bar(2).3) = 2 xx 10^(-2) N`.
Strength of `KOH (gL^(-1))`
`= N xx Eq` or M `xx` Mw = (Mw = Ew of `KOH = 56 g`)
`= 2 xx 10^(-2) xx 56 gL^(-1)`
`= (2 xx 10^(-2) xx 56 xx 200 mL)/(1000mL) g//200 mL`
`= 0.244g//200 mL`
c. [Mw of `Ca(OH)_(2) = 74g, Ew` of `Ca(OH)_(2) = (74)/(2) = 37g]`
`pH = 13.85, pOH = 14 - 13.85 = 0.15`
`log [overset(Θ)OH] = - 0.15 +1 = bar(1).85`
`[overset(Θ)OH] = "Antilog" (bar(1).85) = 7 xx 10^(-1)N` or `(7 xx 10^(-1))/(2)M`
Strength `(gL^(-1)) = (7 xx 10^(-1)N xx 37) gL^(-1)`
or `((7 xx 10^(-1))/(2) xx 74)gL^(-1)`
`= (7 xx 10^(-1) xx 37 xx 100mL)/(1000mL)g//100 mL`
`=2.59g//100mL`