Question

please answer all parts of the question

Question 13 (18 points) A 25.00mL aliquot of 0.376 M butanoic acid is titrated with 0.436 M sodium hydroxide. a Write an equation for the neutralization reaction that is occurring. (1 point) Calculate the initial pH of the acid solution before the titration begins (3 points) Calculate the volume of base solution that is required to reach the equivalence point (3 points) Calculate the pH at the equivalence point (3 points) Calculate the pH after 8.00 mL of the base has been added (3 points) What is the pH halfway to the equivalence point? (1 point) Calculate the pH after 25.00 mL of the base has been added (3 points) h Choose an indicator that can be used for this titration (1 point)

Answer 1

Part by part explanation of the answers are given in the following images.

(a) Neutralization Reaction crbcry crycooH & Naot cezc thcthcooNa + to Here it is to be noted that a buffer solution is formed by the weak acid and its salt with the strong base till complete me neutralization, of the acid takes place. So, it pH can be calculated using Henderson's equation. pH = pka + log [salt] [acid] Gaven for butancic ā, Ka= 1.5x10-5 pka = 4.82 of concentration titration, pH will be that of the weak acid 0 376 M. pH = - 1 log (ka.c) E-log (1-5x10-540-376) = 2624 C) vel of base required can be calculated as vol of base x strength of base - vol of and strenge of acid Vol of bax * 0.436 = 250 0.376 - Vol of bare - 2500 376 - 21-55 ml - Vot of 0.436 (d) At equivalence point, acid and base will have the same concentration and will be neutralized and the pH will solely depend upon the cry cry cry coona salt in aqueous medium Conc. of the salt cone of either acid or base at equivalence / Total volume 0 376) = 0.0094 x1000 = 0.2 (25 to 376 - 46.56 000 21.55 +25)

Now, the salt undergoes reaction as # On, CH, COOK + OH + Na CM CHCOONa+ +6 Initial final 0-2- We know, Kb = - 6.662 10-10 - 1.58105 = 22 Age, ко: [on] [caz Cr Ch Coon] I ovy C H200, COON 6.2-x (x in the denominator is neglected as bis very small) . 6.66 8 10 10 = x² x = 1.154x16 5 which is [OH-] that determines the pH. pou = -log [OH-] = 4.93 pH = 14-POH = 9.062 (e) After 8 me base added, using Henderson equation, T&KO 636 02-0-263 482--0-263 - 4.556 pH : pka + [(25-8) x 0.3761 point, [salt] = [arcial] (f) At half equivalence pH = pka pH = 4.82 (9) Equivalence is reached upon adding 21.55 me, so 25-aiss = 3.45 ml NOOH is what determines the pH 70-436 m) cof Hue Solution 0.36 pot = -logo.436 ...pH = 13.639 (h) Phenolphthalein has range 8.3 -10 and as the pH at. equivalence point is a, phenolphthalein is a suitable indicator. It turns from colowless to pink.