1.

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

Answer» <html><body><p></p>Solution :a. First find the <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of `HCI` re2uired to reach the eqivalent point. <br/> mEq of base = mEq of HCI <br/> `rArr (<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.5 xx (2)/(5) xx 1) = ((2)/(15)xx1)V_(HCI)` <br/> `rArr V_(HCI) = 7.5mL` <br/> The net volume of the solution at the equivalent point <br/> `= V_("base") +V_(HCI) = 2.5 +7.5 = 10mL` <br/> `rArr ["salt"] = c (2//5 xx 2.5)/(10) = 0.1M` <br/> `ph` of an aqueous solution of such a salt is given by : <br/> `pH = 7 - (1)/(2) (pK_(b) +logC)` <br/> `= 7-(1)/(2) (12 + log 0.1) = 1.5` <br/> `rArr [H^(o+)] = "Antilog" (-1.5)` <br/> `= "Antilog" [(-1-0.5 +1-1) = bar(2).5]` <br/> `= 3.2 xx 10^(-2)M` <br/> or <br/> `[H^(o+)] = 10^(-15) = (1)/(10sqrt(10)) = 3.2 xx 10^(-2)M` (Answer 'c') <br/> Please note that answer (c) of the above solution is incorrect. (Why ?) <br/> Actually, `pH = 7 - (1)/(2) (pK_(b) + logC)` is valid only when <br/> `1-h = 1`. <br/> To <a href="https://interviewquestions.tuteehub.com/tag/check-25817" style="font-weight:bold;" target="_blank" title="Click to know more about CHECK">CHECK</a>, claculate `h` using :<br/> `h = sqrt((K_(h))/(c)) = sqrt((K_(w))/(K_(b)C)) = sqrt((10^(-14))/(10^(-12)xx0.1)) = sqrt(0.1)` <br/> `rArr 1 - sqrt(0.1) != 1` <br/> So, we have to solve from basics as follows: <br/> `{:(B^(oplus),+,H_(2)O,hArr,BOH + ,H^(oplus)),(C-Ch,,,,Ch,Ch):}` <br/> `rArr K_(h) = ([BOH][H^(o+)])/([B^(o+)]) = ((Ch)(Ch))/(C-Ch)` <br/> `K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(10^(-12)) = 10^(-2) = 0.01` <br/> Also, <br/> `K_(h) = (Ch^(2))/((1-h))` (Not neglecting `h` and solving, we get) <br/> `0.01 = (0.1 xxh^(2))/(1-h)` <br/> or `h~~ 0.271` <br/> `[H^(o+)] = Ch = 0.27 xx 0.1 = 0.0271 = 2.71 xx 10^(-2) M` <br/> Hence answer is (d). <br/> Note: In such problem, the answer cna be obtained <a href="https://interviewquestions.tuteehub.com/tag/directly-955045" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTLY">DIRECTLY</a>. 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. <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, `[H^(o+)]` would slightly less than the calculated value. <br/> `:. [H^(o+)]` calculated by using formula `= 3.2 xx 10^(-2)M` <br/> `[H^(o+)]` observed (by not using approximation) `= 2.7 xx 10^(-2)M` <br/> 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))` <br/> `=- log [(2)^(5) x 10^(-3)]` <br/> `=-5 log 2+3 = 1.5` <br/> `pH` (observed) (by not usingapproximation <br/> `=- log (27 xx 10^(-3))` <br/> `=- 3log 3+3 = 1.56`</body></html>


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