InterviewSolution
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InterviewSolution
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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`? |
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Answer» 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` |
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