Question

PLEASE SHOW ALL WORK AND STEPS

What is the pH of a solution that contains 0.050 M acetic acid, HC2H3O2 (Ka=1.8 x 10^-5), and 0.05 M hydrocyanic acid HCN (Ka = 4.9 x 1^-10)?

Answer 1

[CH3COOH]= 0.050 M

Ka = 1.8 E-5

[HCN]= 0.05 M

Ka = 4.9 E-10

Lets find concentration of H3O+ from both given acids.

First from Acetic acid

ICE and reaction

            CH3COOH + H2O -- > CH3COO- (aq) + H3O+

I           0.05                             0                      0

C          -x                                 +x                    +x

E          (0.05-x0                       x                      x

Ka = [CH3COO- ][H3O+]/ [CH3COO]

Lets plug the values

1.8 E-5 = x² / (0.05-x)

Value of ka is small (less than 1.0E-4 ) so we can assume the 5 % approximation and neglect the value of x in bracket

1.8 E-5 = x²/ 0.05

Lets now solve for x

x = 9.49 E-4

[H3O+] = x = 9.49 E-4 M

From HCN

            HCN (aq)         +   H2O – > CN^- (aq)    +   H3O⁺ (aq)

I           0.05                                         0                      0

C          -x                                             +x                    +x       

E          (0.05-x)                                               x                      x

Ka = 4.9 E-10 = x² / ( 0.05 –x)

Lets neglect x in the denominator since the ka is very small.

4.9 E-10 = x² / 0.05

Lets calculate value of x

x = 4.95 E-6

[H3O+] = 4.95 E-6 M

Lets assume total solution is 1 L

So molarity of H3O+ from both acetic acid and HCN can be added.

Total molarity of H3O+ = 9.49 E-4 M + 4.95 E-6 = 9.54 E-4 M

Now pH = - log [H3O+]

So pH of this solution = - log ( 9.54 E-4 )

= 3.02

pH of this solution is 3.02