Question

1) The acid-dissociation constant for benzic acid (C6H5COOH) is 6.3 x 10^-5. Calculate the concentrations of H3O+, C6H5COO-, and C6H5COOH in the solution if the initial concentration of C6H5COOH is 0.050 M.

2) Citric acid, which is present in citrus fruit, is a triprotic acid. (a) Calculate the pH of a 0.040 M solution of citric acid. (b) Did you have to make approximations or assumptions in completing your calculations? (c) Is the concentration of citrate ion (C6H5O7^3-) equal to, less than, or greater than the H+ ion concentration?

3) Calculate the molar concentration of OH- in a 0.075 M solution of ethylamine (C2H5NH2; Kb= 4.0 x 10^-6). What is the pH of this solution?

4) Using data from Appendix D, calculate [OH-] and pH for each of the following solution. (a) 0.10 M NaBrO

Answer 1

1)

Ka = 6.3x10^-5

we know that benzoic acid will dissociate to the benzoate ion and H+ in equal concentrations
therefore, [C6H5COO-] = [H+] = x
C6H5COOH --> C6H5COO- + H+
6.3x10^-5 = [C6H5COO-][H+] / [C6H5COOH]
6.3x10^-5 = x^2 / 0.05 - x
3.15x10^-6 - 6.3x10^-5x = x^2
x^2 + 6.3x10^-5x - 3.15x10^-6 = 0
x = 0.0043
[C6H5COO-] = [H+] = 0.0043M
[C6H5COOH] = 0.0457M

2)

Ka1 = [H+] [H2C6H5O7] / [H3C6H5O7]

8.4e-4 = [x] [x] / [0.040 -x]

8.4e-4 [0.040 -x]= x2

x = [H+] = 0.00539 molar =

pH = 2.268 which rounds off to your 2.67.

b)

Here we are just taking the experimental value of Ka as

Citric acid (H3C6H5O7) is a triprotic acid with Ka1 = 8.4 10-4, Ka2 = 1.8 10-5, and Ka3 = 4.0 10-6. Calculate the pH of 0.34 M citric acid and i have done the calculation by taki9ng the Ka1

c)

Yes the conc. of citrate ion is greater than the H+ ion

3)

First write-out the equilibrium equation:

EtNH2 + H2O ---> EtNH3+(aq) + OH-(aq)

Thus the equilibrium expression is given by:

Kb = [EtNH3+][OH-]/[EtNH2]

The concentrations are equilibrium are:

[EtNH3+] = x
[OH-] = x
[EtNH2] = 0.075 - x

Plugging it into the equilibrium equation yields:

(4.0x10^-4) = (x^2)/(0.075-x)

Rearranging gives:

x^2 + (4.0x10^-4)x - (4.8x10^-5) = 0

Using the quadratic equation to solve for x yields a final value of x = 6.73x10^-3, which equals the hydroxide concentration.

SO pOH = 2.171

=> pH = 14 - 2.171 =11.829