Question

I am having trouble understanding this problem. Can you please walk me through the steps using the BCA and ICE tables?

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.68-M HCl to 300. mL of each of the following solutions Change is defined as final minus initial, so if the pH drops upon mixing the change is negative a) water pH before mixing7.00 pH after mixing 0.93 pH change- 6.07 b) 0.158 M C2H3021 pH before mixing 8.97 pH after mixing4.48 pH change- c) 0.158 M HC2H302 pH before mixing 2.77 pH after mixing0.93 pH change - 1.84 d) a buffer solution that is 0.158 M in each C2H32 and HC2H302 pH before mixing- pH after mixing pH change -

Answer 1

b) Consider the ionization of C₂H₃O₂^1- as below.

C₂H₃O₂^1- (aq) + H₂O (l) --------> HC₂H₃O₂ (aq) + OH^- (aq)

Since OH^- is formed, we work with K_a of C₂H₃O₂^1-. We know that K_a (HC₂H₃O₂) = 1.8*10^-5. We further know that

K_a*K_b = K_w

where K_b is the base ionization constant of C₂H₃O₂^1- and K_w = 1.0*10^-14 is a constant. Therefore,

K_b = K_w/K_a = (1.0*10^-14)/(1.8*10^-5)

= 5.55*10^-10

Next, write down the expression for K_b as

K_b = [HC₂H₃O₂][OH^-]/[C₂H₃O₂^1-] = (x).(x)/(0.158 – x)

Since K_b is small, we can assume x << 0.158 M and thus,

K_b = 5.55*10^-10 = x²/(0.158)

====> x² = 5.55*10^-10*0.158 = 8.769*10^-11

====> x = 9.364*10^-6

Therefore, [OH^-] = 9.364*10^-6 M and pOH = -log [OH^-] = -log (9.364*10^-6 M) = 5.028.

We know that pH + pOH = 14; therefore,

pH = 14 – pOH = 14 – 5.028 = 8.972 ? 8.97.

This denotes the initial pH of the solution.

Now, millimoles C₂H₃O₂^1- = (volume of C₂H₃O₂^1- in mL)*(molarity of C₂H₃O₂^1-) = (300. mL)*(0.158 M) = 47.4 mmole.

Millimoles of HCl added = (volume of HCl in mL)*(molarity of HCl) = (10.0 mL)*(3.68 M) = 36.8 mmole.

C₂H₃O₂^1- reacts with HCl as per the equation

C₂H₃O₂^1- (aq) + HCl (aq) --------> HC₂H₃O₂ (aq) + Cl^- (aq)

Cl^- is the counter ion. C₂H₃O₂^1- and HC₂H₃O₂ form a buffer.

As per the stoichiometry of the reaction,

millimoles C₂H₃O₂^1- neutralized = millimoles HCl added = millimoles HC₂H₃O₂ formed = 36.8 mmole.

Millimoles of C₂H₃O₂^1- retained at equilibrium = (47.4 – 36.8) mmole = 10.6 mmole.

Since the total volume of the solution remains stay, we can express the ratio of the concentrations of C₂H₃O₂^1- and HC₂H₃O₂ in terms of the number of moles. We determine the pH using the Henderson-Hasslebach equation as

pH = pK_a + log [C₂H₃O₂^1-]/[HC₂H₃O₂]

====> pH = -log (K_a) + log (moles C₂H₃O₂^1- at equilibrium)/(moles HC₂H₃O₂ at equilibrium)

====> pH = -log (1.8*10^-5) + log (10.6 mmole)/(36.8 mmole)

====> pH = 4.745 + log (0.288)

====> pH = 4.745 + (-0.541)

====> pH = 4.204 ? 4.20

This denotes the pH of the solution after the addition of HCl. Therefore, pH change = (final pH) – (initial pH) = (4.20) – (8.97) = -4.77 (ans).

d) We determine the pH using the Henderson-Hasslebach equation as

pH = pK_a + log [C₂H₃O₂^1-]/[HC₂H₃O₂]

====> pH = -log (1.8*10^-5) + log (0.158 M)/(0.158 M)

====> pH = 4.745 + log (1)

====> pH = 4.745 + 0 ? 4.74

This denotes the pH of the solution before the addition of HCl.

Now, millimoles C₂H₃O₂^1- = millimoles of HC₂H₃O₂ = (volume of C₂H₃O₂^1-/HC₂H₃O₂ in mL)*(molarity of C₂H₃O₂^1-/HC₂H₃O₂) = (300. mL)*(0.158 M) = 47.4 mmole.

Millimoles of HCl added = (volume of HCl in mL)*(molarity of HCl) = (10.0 mL)*(3.68 M) = 36.8 mmole.

C₂H₃O₂^1- reacts with HCl as per the equation

C₂H₃O₂^1- (aq) + HCl (aq) --------> HC₂H₃O₂ (aq) + Cl^- (aq)

Cl^- is the counter ion. C₂H₃O₂^1- and HC₂H₃O₂ form a buffer.

As per the stoichiometry of the reaction,

millimoles C₂H₃O₂^1- neutralized = millimoles HCl added = millimoles HC₂H₃O₂ formed = 36.8 mmole.

Millimoles of C₂H₃O₂^1- retained at equilibrium = (47.4 – 36.8) mmole = 10.6 mmole.

Millimoles of HC₂H₃O₂ at equilibrium = (47.4 + 36.8) mmole = 84.2 mmole.

Since the total volume of the solution remains stay, we can express the ratio of the concentrations of C₂H₃O₂^1- and HC₂H₃O₂ in terms of the number of moles. We determine the pH using the Henderson-Hasslebach equation as

pH = pK_a + log [C₂H₃O₂^1-]/[HC₂H₃O₂]

====> pH = 4.745 + log (moles C₂H₃O₂^1- at equilibrium)/(moles HC₂H₃O₂ at equilibrium)

====> pH = 4.745 + log (10.6 mmole)/(84.2 mmole)

====> pH = 4.745 + log (0.126)

====> pH = 4.745 + (-0.8996)

====> pH = 3.8454 ? 3.84

This denotes the pH of the solution after the addition of HCl. Therefore, pH change = (final pH) – (initial pH) = (3.84) – (4.74) = -0.90 (ans).