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Calculate the pH of the following mixtires of strong acids, strong bases, and combination of both: a. 500mL of 0.1M HCI + 200 mL of 0.1 M H_(2)SO_(4) + 300 mL of 0.2 M HNO_(3) b. 100mL of 0.1 M HCI +100 mL of 0.2 M H_(2)SO_(4) + 100 mL of 0.1 M HNO_(3) and 700 mL of H_(2)O c. 500mL of 0.1 M NaOH +100 mL of 0.1 M Ca(OH)_(2) +400 mL of 0.2 M KOH d. 100mL of 0.1 M Na OH +200mL of 0.1 N Ca (OH)_(2) +200mL of 0.1 M KOH and 500mL of H_(2)O e. 100mL of 0.1M HCI +300mL of 0.1 M H_(2)SO_(4) +100 mL of 0.3 M Ba (OH)_(2) and volume was made to 1L by adding water f 500 mL of 0.1 M HCI +100 mL of 0.1 N H_(2)SO_(4)+ 400 mL of 0.1 M Ca (OH)_(2) g 8g of NaOH +680mL of M HCI +10 mL of H_(2)SO_(4), (specific gravity 1.2, 49% H_(2)SO_(4) bu mass). The total volume of the solution was made to 1L with water. h. 37.0 g of Ca(OH)_(2) +360mL of 1M HCI +10 mL of H_(2)SO_(4) (density = 1.4, 49% H_(2)SO_(4) by mass). The total volume of the solution was made to 1L with water. |
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Answer» Solution :a. All are strong acid and completely ionised. Total volume `= 500 +200 +300 = 100 mL` `N_(1)V_(1) +N_(2)V_(2) +N_(3)V_(3) +N_(4)V_(4)` `N_(4) = (N_(1)V_(1) +N_(2)V_(2) +N_(3)V_(3))/((V_(1)+V_(2)+V_(3))` `(V_(4) = V_(1) +V_(2) +V_(3) = 1000mL)` `0.1 xx 1 xx 500 xx 0.1(n` factor) `xx 200` `N_(A) = (+0.2 xx 1xxx 300)/(1000)` `= (50 +40 +60)/(1000) = (150)/(1000) = 15 xx 10^(-2)N` `PH =- log [H_(3)O^(o+)]` `=- log (15 xx 10^(-2))` `=- log (5 xx3) - log (10^(-2))` `=- 0.7 - 0.48 +2 = 0.82` b. Total volume `= 100 +100 +100 +700 mL H_(2)O` `= 1000mL` `N_(A)=(100xx0.1xx1+100xx0.2xx2(n"factor")+100xx0.1xx1)/(1000)` `= (10+40 +10)/(1000) = (60)/(1000) = 6 xx 10^(-2)N` `:. pH =- log (6 xx 10^(-2)) =- log (3xx2) - log 10^(-2)` `=- log 3 - log 2 +2` `= - 0.48 - 0.3 +2 = 1.22` c. All are strong bases and completely ionised. Thus first claculate `[overset(Θ)OH]` and `pOH` than calculate `pH` accordingly `N_(4)=(500 xx 0.1 xx 1 +100 xx 0.1 xx 2(n "factor") +400 xx 0.2 xx 1)/(500+100+400)` `= (50+20+80)/(1000) = (150)/(1000) = 15 xx 10^(-2) N` `:. pOH =- log [overset(Θ)OH] =- log (15 xx 10^(-2))` `=- log (5 xx 3) - log 10^(-2)` ` =- log 5 - log 3+2` `=- 0.7 - 0.48 +2` `= 0.82` `pH = 14 - 0.82 = 13.18` d. Note: Solution of `CA(OH)_(2)` is given in normally so NEED not to be multiplied by 'n factor' `N_(4) = (100 xx 0.1 xx 1 +200 xx 0.1 xx1 +200 xx 0.1 xx)/(100+200+200 +500 "mL of" H_(2)O)` `= (10 +20 +20)/(1000) = (50)/(1000) = 5 xx 10^(-2)N` `:. pOH =- log [overset(Θ)OH] =- log (5 xx 10^(-2))` `=- log 5 - log 10^(-2)` `=- 0.7 +2 = 1.3` `:. pH = 14 - 1.3 = 12.7` e. Here, mixture of strong acid and strong bases is given. So first calculate th `mEq` of acids and bases. i. IF the `mEq` of strong acid is in excess, then calculate the `pH` from the concentration of `[H^(o+)]` left. ii. If the `mEq` of strong bases is in excess, then calculate the `pOH` from the concentration of `[overset(Θ)OH]` left and then calculate `pH` accrodingly. Total `mEq` of strong acids `= 100 xx 0.1 xx 1 +300 xx 0.1 xx 1("n factor")` `= 10 +60 = 70 mEq` Total `mEq` strong base `= 100 xx 0.3 xx 2("n factor")` `= 60 mEq` `mEq` of strong acid left `= 70 - 60 = 10 mEq` Total volume of solution `= 1L = 1000 mL` `:. [H^(o+)] = (10)/(1000) =10^(-2)N` `pH =- log (10^(-2)) = 2` f. Proceed as in part (e). (Here `H_(2)So_(4)` is given in normality) Total `mEq` of strong acid `= 500 xx 0.1 xx 1 +100 xx 0.1 xx1` `= 50 +10 =60 mEq` Total `mEq` of storng base `= 400 xx 0.1 xx 2("n factor")` `= 8- mEq` `mEq` of strong base left =` 80 - 60 = 20 mEq` Total volume of solution `= 500 +100 +400` `1000 mL` `:.[overset(Θ)OH] = (20)/(100) = 2 xx 10^(-2)N` `:. pOh = - log (2xx 10^(-2))` `=- log2 - log 10^(-2)` `=- 0.3 +2 = 1.7` Thus, `pH = 14- 1.7 = 12.3`. g. Moles of `NaOH = (8)/(40) (Mw of NaOH = 40)` `0.2 mol = 200 m mol` `N_(H_(2)SO_(4)=(%"by weight" xx 10 xxd)/(Ew_(2))` `= (49 xx 10 xx 1.2)/(49) = 12N` `mEq` of `H_(2)SO_(4) = 12N xx 10mL = 120 mEq` `mEq` of `HCI = 680 xx 1 xx 1 = 680 mEq` Total `mEq` of acid `= 120 +680 = 800 mEq` `mEq` of acid left `= 800 - 200 = 600 mEq` Total volume `= 1000 mL` `:. [H^(o+)] = (600)/(1000) = 6 xx 10^(-1)N` `pH =- log (6 xx 10^(-1)) =- log (3xx2) - log 10^(-1)` `=- log 3 - log2 +1` `=- 0.48 - 0.3 +1 = 0.22` h. Mw of `Ca(OH)_(2) = 74` Ew of `Ca (OH)_(2) = (74)/(2) = 37 g Eq^(-1)` eq of `ca (OH)_(2) = (37)/(37) = 1 Eq = 1000 mEq` `N_(H_(2)SO_(4)) = (% "by weight" xx 10 xx d)/(Ew_(2)) = (49 xx 10 xx 1.4)/(49)` `= 14N` `mEq` of `H_(2)SO_(4) = 14N xx 10mL = 140meq` `mEq` of `HCI = 360 mL xx 1N = 360 mEq` Total `mEq` of acid`= (140 +360) = 500 mEq` `mEq` of base left `= 100 - 500 = 500 mEq` Total volume `= 1000 mL` `:. [overset(Θ)OH] (500)/(1000) = 0.5N = 5 xx 10^(-1)N` `pOH =- log [overset(Θ)OH]` `=- log (5 xx 10^(-1))` `=- log 5 - log 10^(-1) =- 0.7 +1 = 0.3` Thus, `pH = 14 - 0.3 = 13.7` |
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