CHEM:ATOM FOC 2E CL (TEXT)
CHEM:ATOM FOC 2E CL (TEXT)
2nd Edition
ISBN: 9780393284218
Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss
Publisher: WW Norton & Co
Question
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Chapter 15, Problem 15.105QA
Interpretation Introduction

To find:

The pKa of 0.125(M) aqueous solution of sodium propanoate

Expert Solution & Answer
Check Mark

Answer to Problem 15.105QA

Solution:

The pKa for 0.125M aquoues solution of sodium propionate is 8.97

Explanation of Solution

1) Concept:

Sodium propionate is a salt of strong base NaOH and weak acid CH3COOH.So it is a basic salt.Na+ ion has no role in its acid-base properties, they are spectator ions.

First, we have to find out the value of Ka from the given value of pKa.

Then, molar concentration of OH- ion has to be calculated using the RICE table for the hydrolysis of Sodium propanoate salt.

pOH is calculated using the molar concentration of OH-

And finally, we calculate pH from the pOH value.

2) Formula:

i) pKa=-logKa

ii) Kh=KwKa

iii) pOH=-logOH-

iv) pH= 14.00-pOH

3) Calculation:

Ka can be calculated as:

4.85=-logKa

Ka=1.41×10-5

Kh=KwKa =1.0×10-141.41×10-5=0.709×10-9

Now we have to calculate the molar concentration of CH3CH2COONa using RICE table:

Reaction:                CH3CH2COO-aq+H2Ol             CH3CH2COOH+OH-

Initial                                          0.125                                                    0                   0

Change                                        -x                                                    +x                  +x

Equilibrium                               (0.125-x)                                            x                   x

To solve for ‘x’, a simplifying assumption has to be made that ‘x’ is very small compared to 0.125M because CH3CH2COO-initial is 0.1250.709 ×10-9 = 0.176 ×109 times than Kb value, which is much greater than 500xguideline.

Kh=CH3CH2COOH[OH-]CH3CH2COO- = xx0.125-xx20.125=0.709×10-9

x=9.414×10-6

So, OH-=9.414×10-6(M)

Therefore, pOH=-log[9.414×10-6]=6-0.9737= 5.026

pH =14.00-5.026=8.97

Conclusion:

The pH of 0.125(M) aqueous solution of Sodium propanoateis 8.97

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Chapter 15 Solutions

CHEM:ATOM FOC 2E CL (TEXT)

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