Question

For problems 1 and 2, assume an activity coefficient of 1 for all substances and no effect of ionic strength. Eliminate terms in quadratic solutions for [H+] only if the weak acid is dissociated < 5%. Reported pKa values can vary depending on the conditions under which they were measured; therefore, in solving the following problems use the pKa values given with the problems.

1. a. What mM concentration of HBr gives a pH of 1.3?

What is the pH of 10 mM NaOH?

At equilibrium 0.1 M nitrous acid (HNO2) produces 7.4 mM NO2-. Calculate the pKa of HNO2.

b. What is the pH of 0.5 M benzoic acid? Benzoic acid Ka = 6.46 x 10-5 M

c. What is the pH of 45 mM H3PO4?

For c-f: Phosphoric acid (H3PO4) is a triprotic acid; pKa1 = 2.12, pKa2 = 7.21, pKa3 = 12.32

Acetic acid pKa = 4.75

d. What is the ratio of H2PO4- to H3PO4 at a pH of 3?

e. If 2 volumes of 18 mM KOH are mixed with 1 volume of 22.5 mM H3PO4, what will be the pH of the final mixture?

f. If 15 μmoles of acetic acid is generated in a 1.0 ml enzymatic reaction buffered by 50 mM Na-phosphate (pH 7.0), what will be the final pH of the reaction mixture?

Would the change in pH be smaller or larger if the reaction were buffered by 50 mM Na phosphate (pH 6.0)? Explain you answer; no calculations required.

2. a. Parietal cells control the concentration of HCl in the stomach by secreting H+ and Cl-. While fasting, they maintain stomach pH at 3.0. After a meal, they decrease stomach pH to 1.4. What is the fold-change in [H+] from fasting to after a meal?

b. For every H+ ion secreted into the stomach, parietal cells also secrete a HCO3- ion into blood plasma. HCO3- plays an important role in maintaining the pH of blood plasma (see textbook page 45 for equations). Patients with achlorhydria have impaired HCl secretion and thus have a loss of HCO3- in the blood. What will be the blood pH of a achlorhydria patient with half the normal concentration of HCO3- in the blood? Assume a constant CO2(d) of 1.2 mM and a normal HCO3- of 24 mM.

c. The pH of the stomach can influence the ionization of drugs such as Aspirin (aka acetylsalicylic acid or HAsp). Given that at equilibrium the pH of a 0.1 M Aspirin is 2.24, what is the pKa of Aspirin?

d. Using the pKa from c., what is the ratio of Asp- to HAsp in the stomach of the patient after a meal?

Would the ratio of Asp to HAsp increase, decrease or stay the same if the patient had fasted?

Answer 1

Answer – 1.a) We are given the pH = 1.3

We know,

pH = -log [H⁺]

so, [H⁺] = 10^-pH

              = 0.0501

We know the HBr is strong acid

So, [H⁺] = [HBr] = 0.0501 M

                             = 50.1 mM

Given, [NaOH] = 10 mM

We know NaOH is strong base

[NaOH] = [OH^-] = 10 mM = 0.010 M

So,

pOH = -log [OH^-]

      = - log 0.010 M

      = 2

So, pH = 14 – pOH

             = 14-2

             = 12

At equilibrium, [HNO₂] = 0.1 M , [NO₂^-] = x = 7.4 mM = 0.0704 M

We know the

Ka = [NO₂^-][H₃O⁺] / [HNO₂]

       = 0.0074 M * 0.0074 M / 0.1 M

       = 5.48*10^-4

So, pKa = -log Ka

              = - log 5.48*10^-4

              = 3.26

Given, [C₆H₅COOH] = 0.5 M

Benzoic acid Ka = 6.46 x 10^-5 M

We need to put ICE table

    C₆H₅COOH + H₂O <----> H₃O⁺ + C₆H₅COO^-

I   0.5                                      0            0

C   -x                                   + x           +x

E 0.5-x                                +x            +x

So, Ka = [H₃O⁺] [C₆H₅COO^-] / [C₆H₅COOH]

6.46 x 10^-5 = x*x/0.5-x

So, x in the 0.5-x can br neglected, since weak acid is dissociated < 5%.

So, x² = 6.46 x 10^-5*0.5

x = 0.00568 M

x = [H₃O⁺] = 0.00568 M

so, pH = -log [H₃O⁺]

          = -log 0.00568 M

          = 2.24