The following questions involve a hypothetical weak acid (HB) (Given the equatio

Question

The following questions involve a hypothetical weak acid (HB) (Given the equation below for the dissociation of this acid in water Identify the Bronsted acids and bases in the reaction HB(aq) + H_2O rightarrow H_3 O*(aq)^+ B (aq) Determine the K_4 of this acid if a 0.50M solution has a pH of 2.75. If 50.0ml of a 0.50M weak acid is titrated with 1.00M NaOH. What is the pH after 10.0ml of the base is added? What is the pH of the solution at the equivalence point of the titration described in C above? Use lessthanorequalto Chatelier's principle to explain how the addition of the OH ions first creates a buffer solution and then a weak base solution at the equivalence point.

Explanation / Answer

Question 2.

a)

First, let us define Bronsted Lowry acid/base:

Bronsted Lowry acid: any species that will donate H+ (protons) in solution, and makes pH lower (i.e HCl)

Bronsted Lowry base: any species that will accept H+ (protons) in solution, and makes pH higher (NH3 will accept H+ to form NH4+)

Typically, acid/bases are shown in the left (reactants)

when we write the products:

Bronsted Lowery conjugate base = the base formed when the B.L. acid donates its H+ proton ( i.e. HCl -> Cl-

Bronsted Lowery conjugate acid = the acid formed when the B.L. base accept its H+ proton ( i.e. NH4+ has accept H+ proton)

Note that, typically conjugate bases/acids are shown in the right (product) side

then, for

HB + H2O = H3O+ + B-

bronsted acid = HB

bronsted base = H2O

bronsted conjugate base = B-

bronsted conjugate acid = H3O+

b)

Ka for this acid if:

Ka = [H+][A-]/[HA]

[H+] = [A-] = 10^-pH = 10^-2.75 = 0.0017782

[HA ] = M-x = 0.5 -x = 0.5-0.0017782 = 0.4982218 M

so

Ka = (0.0017782*0.0017782)/0.4982218 = 6.346*10^-6

c)

mmol of acid = MV = 50*0.5 = 25 mmol

mmol of base = MV = 10*1 = 10

mmol of acid left = 25-10 = 15

conjugate formed = 0+10 = 10

substitte in pH equation

pH = pKa + log(A-/HA) = -log( 6.346*10^-6) + log(10/15)

ph = 5.20 + log(10/15) = 5.023

d)

in equivalence point

V = Vacid + Vbase

Vacid = 50

Vbase = (50*0.5)/(1) =25

Vtota = 75 mL

[B-] = 25/75 = 0.333

Kb = [HB][OH-]/[B-]

(10^-14)/(6.346*10^-6) = x^x/(0.33-x)

1.575*10^-9 = x*x/(0.33-x)

x = 2.29*10^-5

pOH = -log(2.29*10^-5) = 4.6401

pH = 14-4.6401 = 9.3599

e

addition of OH- will be buffered since HB is present so H2O is formed and B- is formed as well

in the equivalence point, there is no conjguate which will react with OH- so OH- addition will change the pOH drastically

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