(c) Aqueous aniline, CsH,NH2 reacts with formic acid, HCOOH. 2. You prepare a bu
ID: 1036712 • Letter: #
Question
(c) Aqueous aniline, CsH,NH2 reacts with formic acid, HCOOH. 2. You prepare a buffer that is 0.35 M Na,HAsO,(sodium hydrogen arsenate) and 0.45 M NaH Aso, (sodium dihydrogen arsenate). (a) The acid component of this buffer is (b) The base component of the buffer is (c) The spectator ion(s) that will never do anything important for this buffer is/are Calculate the pH of this solution. (d) (e) Write the reaction that occurs when HCl (or H) is added to this buffer (1) Write the reaction that occurs when NaOH (or OH) is added to this buffer. (g) If 1.75 mL of 1.50 M HCl is added to 125.0 mL of this buffer, what is the pH of the resulting solution? Acid Dissociation Constants for Select Acids at 25°C 30 boExplanation / Answer
Ans. #a. NaH2AsO4 or H2AsO42- acts as the acid component.
#b. Na2HAsO4 or HAsO43- acts as the base component (conjugate base).
#c. Na+ ions are spectator ions.
#d. pKa = -log Ka of H2AsO42- = -log 9.3 x 10-8 = 7.03
Using HH equation: pH = pKa + log ([A-] / [AH])
Where, AH = weak acid, NaH2AsO4
A- = Conjugate acid Na2HAsO4
Putting the values in above expression-
pH = 7.03 + log (0.35 / 0.45) = 7.03 + (-0.11) = 6.92
#e. Addition of HCl: H+ reacts with conjugate base HAsO43- for form the acid as follow- HAsO43- + H+ ------> H2AsO42-
#f. Addition of NaOH: OH- reacts with the acid to form the conjugate base as follow- H2AsO42- + OH- ------> HAsO43-
#g. Concertation after mixing HCl to the Buffer:
Total volume = 125.0 mL + 1.75 mL = 126.75 mL
Now, using C1V1 (original soln.) = C2V2 (mixed soln.) for the mixed solution-
[H2AsO42-] = (C1V1 / C2) = 0.45 M x 125.0 mL / 126.75 mL = 0.4438 M
[HAsO43-] = (C1V1 / C2) = 0.35 M x 125.0 mL / 126.75 mL = 0.3452 M
[HCl] = 1.50 M x 1.75 mL / 126.75 mL = 0.0207 M
# Being strong acid, HCl undergoes complete dissociation in aqueous solution. 1 mol H+ from HCl reacts with 1 mol HAsO43- to form 1 mol H2AsO42-.
So,
[HAsO43-] neutralized = [H+] from HCl = 0.0207 M
[H2AsO42-] formed = [HAsO43-] neutralized = 0.0207 M
Total [H2AsO42-] = Initial [H2AsO42-] + [H2AsO42-] formed
= 0.4438 M + 0.0207 M = 0.4645 M
Remaining [HAsO43-] at equilibrium = Initial [HAsO43-] – [HAsO43-] neutralized
= 0.3452 M – 0.0207 M = 0.3245 M
# Again, Using HH equation –
pH = 7.03 + log (0.3245 / 0.4645) = 7.03 + (-0.16) = 6.87
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