Question

At a temperature of 25.0◦C, we dissolve 3.33 g of an acid, HA, in enough water to produce 25.0 mL of solution. We titrate this solution with a 0.444 M NaOH solution and we require 18.8 mL to reach the equivalence point. The ionization constant of HA, Ka, is 5.5 × 10−6 . The temperature is 25.0◦C throughout. (a) (2 points) What is the molar mass of HA? (b) (2 points) What was the original pH of the HA(aq) solution? (c) (2 points) What is the pH at the equivalence point? (c) (2 points) What was the pH after the addition of 9.4 mL of the required 18.8 mL of the 0.444 M NaOH solution (so exactly at the midpoint of the titration where we had equal amounts of HA(aq) and A−(aq))? Answer: a) 399 b) 2.87 c) 9.27 d) 5.26

Answer 1

Please see the step by step solution attached below.

Parts a) and b)

At = 10 25C, kw (for water) pkw=14 HA (aq) + Naon (aq) NaA Laq) + 4₂0 (1) ИА and Naon react in 1) molar ratio mmol of Naoh added at equivalence (0.444M)*(18.8 mL) 8.35 point = mmol of HA present ИА Amount of 10 3-33g Molar mass HA of = 3.33 g 8.35 mmol 3.33 8.35x103 mo) - 398.8 g/mol 1 399 g/mol 3 significant digits Answer a) upto 11 b) [HA] 8.35 mmo) 25.0mL mmol of HA volume of solution (inme) 10.334 mol/L ka initially: HA (aq) + H+ (aq) + A-laq) x -6 i Ka=5.5x10 x is small M: (0.334 - x) very less (mol) [ut] [A x² ka x² 10.334) & [HA] (0-334- 2= 1-35x10-3 0.334 x 5.5x100 [u+ м M and 1:35x103 log [ut] = -log (1-35x103) pH = - log 3- log (1:35) 3 -0.130 = 2.869 164 here, initial pH = 2.87 Answer

Part c)

At equivalence point due to hydrolysis of A- (conjugate base of HA) , excess OH- will be present.

See attached calculations:

c) At the equivalence point, au and been neutralized converted has of HA to NA . Na A nmol A mmol of of initial mmol of HA 8.35 Total volume - solution of (2500 mL) + (18.8 mL) = 43.8mL [A-] = 0.191 mol/L = 10.191 M 8.35 mmol 43.8 mL now, a weak A undergo hydrolysis as: (: it is conjugate base of aid) Alaq) + H₂0 (1) HA (aq) +oh Lag) (0:191-2') u' a' A (+) 9 Kn= kw -14 - 10 1.82 x 10 [ou] [HA [A- ka 5:58106 or, 1.82x10 9 x2 2012 (sine, i' will be 0191 (0-191-! small) Y2 * a' (1.82x109 x0.191) = 1.86x105 1.86x105 m Cour] = x' M = and POM - - log [or] = 5- log (1•8 6) = 4073 and pH + pon pH= 14 14- pOH = 14-4.73=9.27 pH=9.27 So, Answer

Part d)

at half the equivalence point, half of HA (acid) has been neutralized and converted to A- (conjugate base) ,

and [HA] = [A-]

this will form a buffer solution and pH is given as:

pH = pka +log [con jugatebase] [acid]

The log term is zero, as log 1 = 0,

so, pH = pKa = -log (5.5*10^-6) = 6 - log 5.5 = 5.26   ....Answer

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