Question

1. In a titration of butanoic acid with NaOH, calculate the mols of butanoic acid, BuH,and mols of butanoate, Bu-, after mixing 10,000mL of 0.080 M butanoic acid, with 2.00mL of NaOH 0.01M.         

BuH (aq) + OH^- (aq)  à Bu- (aq) + H₂O (l)

2. What is the proton concentration of the butanoic acid buffer (Ka=1.5×10⁻⁵) after the addition of NaOH in A)

SHOW WORK

Answer 1

Sol: Titration of butanoic acid with HapH NaOH is a 0.08 I x 10 L Butanoic acid (BUH) is a weak acid and strong base. The values provided in the question are, Volume of Buh= 10,000 me = lot Concentration of Buh= 0.08 M = 0.08 mod Mole of BuH = -0.8 mol Volume of NaOH = 2.00 me = 2.00 x103 L concentration of NaoH = 0.01 M = 0.01 mod Mole of NaOH 2.00 x103 L x 0.01 = 2x10 5 mol - o.ooooa mot BUH (aa) OH (aa) Buh (and 0.8mol 0.00002 mol mal L + H₂O Start Final 0.7998 mol 0.00002 mol After completion of reaction, whole Oh will be consumed. So, after mixing, mole of BuH 0.7998 mol Mole of Butanoate & = 2x105 mol

After mixing of But and NaOH, it forms buffer solution. Because there are salt of weak acid (Buna) and weak Acid (BUH), in the solution. So. According to Henderson - Hasselbalch equation, pH = pka + pka + logo log. [BuH] [Bu Ka -log, [Bu] logo [BUH) lo = -log, (1.5x105) [2x105 4 or - log [ht] 0r, -dog. Ing logo 4.8 10 [o. 7998] 4.60 Or - log Tut] 0.2 OY, lo = 0.63 Or, [H So, Proton concentration of butanoic acid buffer after the addition of NaOH. is 0.63 M.

As NaOH is strong base and it will completely dissociate. So it wii neutralize equal mole of H+ (proton) , which is available in the solution. So final solution will be buffer solution.