Question

Find the pH of each of the following solutions of mixtures of acids.

0.190 M in HCHO2 and 0.225 M in HC2H3O2

Express your answer to two decimal places.

**answers are not 2.10, 2.11, or 2.24**

ka for HCHO2= 1.8E-4

ka for HC2H3O2=1.8E-5

please show work, thank you!

Answer 1

1)

Lets write the dissociation equation of HCHO2

HCHO2 -----> H+ + CHO2-

0.19 0 0

0.19-x x x

Ka = [H+][CHO2-]/[HCHO2]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.8*10^-4)*0.19) = 5.848*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.8*10^-4 = x^2/(0.19-x)

3.42*10^-5 - 1.8*10^-4 *x = x^2

x^2 + 1.8*10^-4 *x-3.42*10^-5 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 1.8*10^-4

c = -3.42*10^-5

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 1.368*10^-4

putting value of d, solution can be written as:

x = {-1.8*10^-4 + √(1.368*10^-4)}/2

x = {-1.8*10^-4 - √(1.368*10^-4)}/2

solutions are :

x = 5.759*10^-3 and x = -5.939*10^-3

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

x = 5.759*10^-3

So, [H+] = x = 5.759*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (5.759*10^-3)

= 2.24

Answer: 2.24

2)

Lets write the dissociation equation of HC2H3O2

HC2H3O2 -----> H+ + C2H3O2-

0.225 0 0

0.225-x x x

Ka = [H+][C2H3O2-]/[HC2H3O2]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.8*10^-5)*0.225) = 2.012*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.012*10^-3 M

So, [H+] = x = 2.012*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (2.012*10^-3)

= 2.70

Answer: 2.70