Question

In: Chemistry

Calculate the [H ] and pH of a 0.000183 M acetic acid solution. Keep in mind...

Calculate the [H ] and pH of a 0.000183 M acetic acid solution. Keep in mind that the Ka of acetic acid is 1.76 × 10^-5.

Solutions

Expert Solution

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

1.83*10^-4 0 0

1.83*10^-4-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.76*10^-5)*1.83*10^-4) = 5.675*10^-5

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.76*10^-5 = x^2/(1.83*10^-4-x)

3.221*10^-9 - 1.76*10^-5 *x = x^2

x^2 + 1.76*10^-5 *x-3.221*10^-9 = 0

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

a = 1

b = 1.76*10^-5

c = -3.221*10^-9

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 = 1.319*10^-8

roots are :

x = 4.863*10^-5 and x = -6.623*10^-5

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

x = 4.863*10^-5

so.[H+] = x = 4.863*10^-5 M

we have below equation to be used

pH = -log [H+]

= -log (4.863*10^-5)

= 4.31

Answer: 4.31


Related Solutions

Calculate the [H ] and pH of a 0.0045 M hydrazoic acid solution. Keep in mind...
Calculate the [H ] and pH of a 0.0045 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20 × 10-5. Use the method of successive approximations in your calculations or the quadratic formula.? [H+] = M? pH = ?
Calculate the [H ] and pH of a 0.000501 M hydrofluoric acid solution. Keep in mind...
Calculate the [H ] and pH of a 0.000501 M hydrofluoric acid solution. Keep in mind that the Ka of hydrofluoric acid is 6.80 × 10-5.
Calculate the pH and pOH of a 0.50 M solution of Acetic Acid. The Ka of...
Calculate the pH and pOH of a 0.50 M solution of Acetic Acid. The Ka of HOAc is 1.8 x10-5 [H3O+]/(0.50-0.00095)=1.8x10-5 where did they get the .00095, its said to take that as the second assumption [H3O+]=9.4x10-4 [H3O+]/(0.50-0.00094)=1.85x10-5 [H3O+]=9.4x10-4 (this is said to be the third assumption) Please, please help I really dont understand this, please show all steps and be as descriptive as possible.
Calculate the ionization constant of acetic acid: pH of 0.01 M acetic acid: 3.50 pH of...
Calculate the ionization constant of acetic acid: pH of 0.01 M acetic acid: 3.50 pH of 1.00 M acetic acid: 2.34
The pH of an aqueous solution of 0.441 M acetic acid is​ _______________
The pH of an aqueous solution of 0.441 M acetic acid is​ _______________
Calculate the percent ionization and the expected intial pH for the 0.10 M acetic acid solution
Calculate the percent ionization and the expected intial pH for the 0.10 M acetic acid solution
Calculate the pH of a solution of 0.25 M Acetic acid and 0.34M Sodium acetate before...
Calculate the pH of a solution of 0.25 M Acetic acid and 0.34M Sodium acetate before and after 0.036 M NaOH is added to the solution. Ka of HAc = 1.8 x 10 -5 Please Use the ICE method
Calculate the pH of a 0.800 M NaCH3CO2 solution. Ka for acetic acid, CH3CO2H, is 1.8...
Calculate the pH of a 0.800 M NaCH3CO2 solution. Ka for acetic acid, CH3CO2H, is 1.8 × 10-5. If someone could list the steps and calculations that would be helpful. Thank you!
Calculate the pH of a 0.100 M sodium benzoate, NaC6H5CO2, solution. The Ka of acetic acid...
Calculate the pH of a 0.100 M sodium benzoate, NaC6H5CO2, solution. The Ka of acetic acid is 6.3 x 10-5
Calculate the pH of a 0.51 M CH3COOK solution. (Ka for acetic acid = 1.8×10−5.)
  Calculate the pH of a 0.51 M CH3COOK solution. (Ka for acetic acid = 1.8×10−5.)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT