Question

What is the pH of a 0.5 m solution of sodium propionate, NaC3H5O2 at 25 degrees...

What is the pH of a 0.5 m solution of sodium propionate, NaC3H5O2 at 25 degrees C?

(For propionic acid, HC3H5O2, Ka = 1.3 x 10-5 at 25 degrees C)

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

The Ka of acid will be

CH3CH2COOH --> CH3COO- + H+

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

The sodium acetate will hydrolyze as

CH3CH2COO-Na+ + H2O --> CH3CHCOOH + Na+ + OH-

Kh = Kwater / Ka

Kh = 10^-14 / 1.3 x 10-5

Kh = 0.769 X 10^-9

Kh = [CH3CH2COOH] [ OH-] / [CH3CH2COO-]

Kh = x^2 / 0.5-x

0.769 X 10^-9 = x^2 / 0.5-x

We may ignore x in denominator as Kh is very low

0.769 X 10^-9 = x^2 / 0.5

x^2 = 0.385 X 10^-9

x= 1.96 X 10^-5

[OH-]= 1.96 X 10^-5

pOH = -log[OH-] = 4.71

pH = 14-4.71 = 9.29

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
What is the pH of a 0.30 M solution of sodium propionate, NaC 3H 5O 2,...
What is the pH of a 0.30 M solution of sodium propionate, NaC 3H 5O 2, at 25°C? (propionic acid, HC 3H 5O 2, is monoprotic and has a Ka = 1.3 × 10 –5 at 25°C.. K w = 1.01 × 10 -14 )
A propionic acid (CH3CH2COOH) buffer is prepared by adding 0.15 moles of propionic acid and 0.10...
A propionic acid (CH3CH2COOH) buffer is prepared by adding 0.15 moles of propionic acid and 0.10 moles of sodium propionate (NaCH3CH2COO) to water to make 1.20 L of solution. The Ka of propionic acid is 1.3 x 10-5.   Then 2.5 mL of 1.00 M KOH is added. What is the new pH?
A solution of 10-3-mole of sodium propionate (CH3CH2COONa) is added to 1 liter of distilled water....
A solution of 10-3-mole of sodium propionate (CH3CH2COONa) is added to 1 liter of distilled water. If the pKa = 4.87 for propionic acid, what the solution pH? Solve the appropriate equations analytically and graphically and compare the results. Also, using the same graph determine the pH if a 10-3-mole of propionic acid (CH3CH2COOH) were added instead of the sodium propionate. Step by step explanations would be appreciated. Thanks!
A. Consider a buffer solution that is 0.30M propionic acid and 0.40M sodium propionate. Ka (propionic...
A. Consider a buffer solution that is 0.30M propionic acid and 0.40M sodium propionate. Ka (propionic acid )=1.3x105 a. Calculate the [H+], {OH-] and pOH for this buffer b. Calculate the pH of this buffer after adding 20ml of 2.0M NaOH to 1.0L of the solution (Hint: 1.0L + 20ml = 1.0L according to the rules of addition for significant figues)
Nitric acid and propionic acid (CH3CH2CO2H) were combined to give 200. mL of an aqueous solution...
Nitric acid and propionic acid (CH3CH2CO2H) were combined to give 200. mL of an aqueous solution with a total solute concentration of 0.200 M (before dissociation of each acid). The pH of the solution (after equilibrium was reached) was 1.50. Ka = 1.3 x 10-5 for propionic acid. Kw = 1.0 x 10-14. What were the equilibrium concentrations of hydrogen ion, nitrate ion, propionate ion, and propionic acid in the solution?
For a solution that is 0.285 M HC3H5O2 (propionic acid, Ka=1.3×10−5) and 0.0892 M HI, calculate...
For a solution that is 0.285 M HC3H5O2 (propionic acid, Ka=1.3×10−5) and 0.0892 M HI, calculate the following. [OH−] [C3H5O−2] [I−]
1. A buffer is made by mixing 1.250 mol formic acid, HCOOH, with 0.750 mol sodium...
1. A buffer is made by mixing 1.250 mol formic acid, HCOOH, with 0.750 mol sodium formate, HCOONa in enough water for a total volume of 1.00L. The Ka of formic acid is 1.87 x 10-4 a. What is the pH of this buffer? b. If 0.075 mol HCl is added to this buffer, what is the pH of the resulting buffer? (assume no volume changes) c. If 0.055 mol NaOH is added to this buffer, what is the pH...
A  25.00-mL  sample of propionic acid,  HC3H5O2,  of unknown concentration A  25.00-mL  sample of propionic acid,  HC3H5O2,  of unknown concentration  was titrated w
A  25.00-mL  sample of propionic acid,  HC3H5O2,  of unknown concentration A  25.00-mL  sample of propionic acid,  HC3H5O2,  of unknown concentration  was titrated with  0.151  M  KOH. The equivalence point was reached when was titrated with  0.151  M  KOH. The equivalence point was reached when  41.28  mL  of base had been added. What is the pH at the equivalence point?    41.28  mL  of base had been added. What is the pH at the equivalence point?     Ka   for propionic acid is    1.3×10–5   at  25°C.Ka   for propionic acid is    1.3×10–5   at  25°C. Select one: a. 8.93 b. 7.65 c. 9.47 d. 5.98 e. 9.11
Calculate the pH at 25°C of a 0.075 M solution of a weak acid that has...
Calculate the pH at 25°C of a 0.075 M solution of a weak acid that has Ka = 1.3 ×10−5.
Calculate the pH at the equivalence point in titrating 0.089 M solutions of each of the...
Calculate the pH at the equivalence point in titrating 0.089 M solutions of each of the following with 0.061 M NaOH. (a) hydrochloric acid (HCl) pH =   (b) propionic acid (HC3H5O2), Ka = 1.3e-05 pH =   (c) phenol (HC6H5O), Ka = 1.3e-10 pH =  
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT