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2.5mLof 2//5M weak mono-acidic base (K_(b) = 1 xx 10^(-12) at 25^(@)C) is titrated with 2//15M HCI in water at 25^(@)C. Find the concentration of H^(o+) ions at equivalence point. (K_(w) = 1 xx 10^(-14)at 25^(@)C) a. 3.7 xx 10^(-13)M b. 3.2 xx 10^(-7)M c. 3.2 xx 10^(-2)M d. 2.7 xx 10^(-2)M |
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Answer» Solution :a. First find the VOLUME of `HCI` re2uired to reach the eqivalent point. mEq of base = mEq of HCI `rArr (2.5 xx (2)/(5) xx 1) = ((2)/(15)xx1)V_(HCI)` `rArr V_(HCI) = 7.5mL` The net volume of the solution at the equivalent point `= V_("base") +V_(HCI) = 2.5 +7.5 = 10mL` `rArr ["salt"] = c (2//5 xx 2.5)/(10) = 0.1M` `ph` of an aqueous solution of such a salt is given by : `pH = 7 - (1)/(2) (pK_(b) +logC)` `= 7-(1)/(2) (12 + log 0.1) = 1.5` `rArr [H^(o+)] = "Antilog" (-1.5)` `= "Antilog" [(-1-0.5 +1-1) = bar(2).5]` `= 3.2 xx 10^(-2)M` or `[H^(o+)] = 10^(-15) = (1)/(10sqrt(10)) = 3.2 xx 10^(-2)M` (Answer 'c') Please note that answer (c) of the above solution is incorrect. (Why ?) Actually, `pH = 7 - (1)/(2) (pK_(b) + logC)` is valid only when `1-h = 1`. To CHECK, claculate `h` using : `h = sqrt((K_(h))/(c)) = sqrt((K_(w))/(K_(b)C)) = sqrt((10^(-14))/(10^(-12)xx0.1)) = sqrt(0.1)` `rArr 1 - sqrt(0.1) != 1` So, we have to solve from basics as follows: `{:(B^(oplus),+,H_(2)O,hArr,BOH + ,H^(oplus)),(C-Ch,,,,Ch,Ch):}` `rArr K_(h) = ([BOH][H^(o+)])/([B^(o+)]) = ((Ch)(Ch))/(C-Ch)` `K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(10^(-12)) = 10^(-2) = 0.01` Also, `K_(h) = (Ch^(2))/((1-h))` (Not neglecting `h` and solving, we get) `0.01 = (0.1 xxh^(2))/(1-h)` or `h~~ 0.271` `[H^(o+)] = Ch = 0.27 xx 0.1 = 0.0271 = 2.71 xx 10^(-2) M` Hence answer is (d). Note: In such problem, the answer cna be obtained DIRECTLY. The answer (c) `(3.2 xx 10^(-2)M)` is calculated by using the formula of salt of `W_(B)//S_(A)`. The concentration of salt is `0.1M` (which is very high). So `(1-h)` cannot be taken equal to `1`, since the two answer (c) and (d) are very close to each other. THEREFORE, `[H^(o+)]` would slightly less than the calculated value. `:. [H^(o+)]` calculated by using formula `= 3.2 xx 10^(-2)M` `[H^(o+)]` observed (by not using approximation) `= 2.7 xx 10^(-2)M` In such problem, if the `pH` is asked, calculated it by using formula will be slightly less than the observed `pH` (by not using approcimation) e.g., `pH` (calculated by using formula) `=- log (3.2 xx 10^(-2))` `=- log [(2)^(5) x 10^(-3)]` `=-5 log 2+3 = 1.5` `pH` (observed) (by not usingapproximation `=- log (27 xx 10^(-3))` `=- 3log 3+3 = 1.56` |
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