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Calculate the pH of solution made by mixing equal volume of : a. Two solutions having pH = 1.5 and 2.5. b. Three solutions having pH = 15,2.5, and 3.5. c. Two solutions having pH = 8 and 9. d. Three solutions having pH 8,9, and 10. e. Two solutions having pH =2and 4. f.Three solutions having pH = 2,4, and 6. |
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Answer» Solution :a. The two `pH` values differe by one unit. Approximate final `pH = ((1.5 +2.5))/(2) = 2` Actural final `pH = 2.0 - 0.24 = 1.76` Checking final pH : `pH = 1.5, log [H^(o+)] =- 1.5 =- 1- 0.5 +1 -1 = bar(2).5` `[H^(o+)] = "Antilog" (bar(2).5) ~~3xx 10^(-2)N` `pH = 2.5, log [H^(o)]=- 2.5 =p- 2- 0.5 +1 -1 = bar(3).5` `[H^(o+)] = "Antilog" (bar(3).5) ~~ 3xx 10^(-3)N` Final `[H^(o+)] = ((3xx10^(-2)+3xx10^(-3)))/(2)` `= (3xx10^(-2))/(2) [1+0.1]` `=(1.1)/(2) xx3xx 10^(-2)` `0.55 xx 3xx 10^(-2) = 55 xx 3xx10^(-4)` `pH =- log 3 - log 10^(-4)` `=- 1.74 - 0.5 +4` (taking log `3~~0.48 ~~0.5`) `=-2.24 +4 = 1.76` b. The three `pH` values differe by one unit Approximate final `= ((1.5+2.5+3.5))/(3) = (7.5)/(3) = 2.5` Actual final `pH = 2.5 - 0.56 = 1.94` Checking final pH : Proceed as above in pair (a) `pH = 1.5, :. [H^(o+)] ~~3xx10^(-2)N` `pH =2.5, :. [H^(o+)] ~~3 xx 10^(-3)N` `pH = 3.5 :. [H^(o+)]~~3 xx 10^(-4)N` Final `[H^(o+)] = (3xx10^(-2)+10^(-3)+10^(-4))/(3)` `=3xx10^(-2) (1+0.1+0.01)` (neglect 0.01) `= (1.1 xx 3)/(3) xx 10^(-2)` `=0.366 xx3x10^(-2) = 366 xx 3xx 10^(-5)` `pH =- log (366) - log3 - log 10^(-5)` `=- 2.5635 - 0.5 +5` (taking log `3 = 0.48 ~~0.5`) `=-3.0635 +5 = 1.9365 ~~1.94` `pH = 1.94` c. Although the `pH` values are in basic range, so `pH` of mixture can be calculated as when `pH` values are in acidic range. Approximate final `pH = ((8+9))/(2) = 8.5` Actual final `pH = 8.5 - 0.24 = 8.26` Checking final pH : `pH = 8:. [H^(o+)] = 10^(-8)N` `pH = 9:. [H^(o+)] = 10^(-9)N` Final `[H^(o+)] = ((10^(-8)+10^(-9))/(2)` `=(10^(-8)(1+0.1))/(2)` `= (1.1)/(2) xx 10^(-8)N` `=0.55 xx 10^(-8)= 55 xx 10^(-10)N` `pH =- log (55) - log10^(-10) =- 1.74 +100 = 8.26`. d. Procedd as in part (c) Approximate final `pH = ((8+9+10))/(3) = 9` Actual final `pH = 9.0 - 0.56 = 8.44` Checking final pH : Final `[H^(o+)] = (10^(-8)+10^(-9)+10^(-10))/(3)` `= (10^(-8)(1+0.1+0.01))/(3)` (neglect 0.01) `= (1.1)/(3) xx 10^(-8) = 0.366 xx 10^(-8) = 366 xx 10^(-11)` `pH =- log (366) - log 10^(-11)` `=- 2.5635 +11 = 8.365 ~~ 8.44` e. The two `pH` values do not differ by one unit, so calulate the `pH` directly `[H^(o+)]_("TOTAL") = ((10^(-2)+10^(-4))/(2)) = (10^(-2))/(2) (1+10^(-2))` (neglect `10^(-2)`) `~~(10^(-2))/(2) ~~0.5 xx 10^(-2)` `~~5xx10^(-3)M` `pH ~~ - log (5x10^(-3)) ~~- 0.7 +3 ~~2.7` f. Here also, the three `pH` VALUE do not differ by one unit, so calculate the `pH` directly. `[H^(o+)]_("total") = ((10^(-2)+10^(-4)+10^(-6))/(3)) = (10^(-2))/(3) (1+10^(-2)+10^(-4))` (neglect `10^(-2)` and `10^(-4)` ) `~~ (10^(-2))/(3)~~ 0.3 xx 10^(-2)` `~~ 3 xx 10^(-3)M` `pH ~~ - log (3xx10^(-3)) ~~ 0.48 +3 ~~ 2.52` |
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