Question

If pKa of weak acids are the following: pKa acetic acid = 4.75 pKa propionic acid = 4.87 pKa acitric acid = 5.40 pKa of benzoic acid = 4.19

What is the order of acids with increasing acidity Determine the pH of the following

A. 100 mL of benzoic acid solution 0.03M is mixed with 200 ml water

B. 50 mL of sodium hydroxide solution 0.05 M is mived with 200 ml water

Answer 1

1)

Greater value of Ka implies stronger acid.

So, lesser pKa value implies stronger acid.

Order of increasing acidity is:

acitric acid< propionic acid < acetic acid < benzoic acid

2)

A)

[C6H5COOH] = mol of C6H5COOH / volume of solution

= M(C6H5COOH)*V(C6H5COOH) / total volume

= 0.03*100 / (200+100)

= 0.010 M

use:

pKa = -log Ka

4.19 = -log Ka

Ka = 6.457*10^-5

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

1*10^-2 0 0

1*10^-2-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.457*10^-5)*1*10^-2) = 8.035*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.457*10^-5 = x^2/(1*10^-2-x)

6.457*10^-7 - 6.457*10^-5 *x = x^2

x^2 + 6.457*10^-5 *x-6.457*10^-7 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 6.457*10^-5

c = -6.457*10^-7

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 2.587*10^-6

roots are :

x = 7.719*10^-4 and x = -8.365*10^-4

since x can't be negative, the possible value of x is

x = 7.719*10^-4

So, [H+] = x = 7.719*10^-4 M

use:

pH = -log [H+]

= -log (7.719*10^-4)

= 3.1124

Answer: 3.11

B)

[NaOH] = mol of NaOH / volume of solution

= M(NaOH)*V(NaOH) / total volume

= 0.05*50 / (200+50)

= 0.010 M

use:

pOH = -log [OH-]

= -log (1*10^-2)

= 2

use:

PH = 14 - pOH

= 14 - 2

= 12

Answer: 12.00