2:32 PM nknow acid H AVERAGE 4. 42 4.84 48 10.1 12.11 5. UNKNOW ACID: 0.201 M Na

Question

2:32 PM nknow acid H AVERAGE 4. 42 4.84 48 10.1 12.11 5. UNKNOW ACID: 0.201 M NaOH 0.1005M Analysis : (A1) Write the chemical equation (A2) Calculate Ka for each solution. At least for one solution, show your work clearly and completely so anyone can follow what you have done to calculate Ka. (A3) Calculate average Ka. (one that would lead to the Ka) and Ka spression Postlab Questions : (Q1) What would happen to the final answer if as your total volume in your calculation? (Q2) The calculations are done assuming that OH-added is reacting completely, What would hacpen t the final result if not all OH- was used up? Hint: WHAT is the final result? What would happen to the VALUE? you forgot to add water to test tube #1 but used 20.00m Why?

Explanation / Answer

Unknown acid 0.201

NaOH = 0.1005

Analysis

A) We can write the chemical equation for the dissociation of a weak acid HA, using A- to represent the conjugate base.

HA <==> A- +H+

The equilibrium expression for the dissociation of a weak acid is written as

Ka =[A-] [H+]/[HA]

The dissociation constant of an acid = to the concentration of hydrogen ions * the concentration of the conjugate base of the acid/ the concentration of un-ionized acid.

B)

Given

Molarity of unknown acid = 0.201M

pH = 4.42

1)Ka when pH is 4.42

HA   <==>           A-   +      H+

I 0.201                 0            0

C –x                     +x          +x

E0.201-x             x            x

Calculate [H+] from pH

pH = log[H+]

4.42 = -log [H+]

[H+] = 10^-pH = 10^ -4.42

[H+] = 3.80 x 10^-5

The equilibrium expression for the dissociation of a weak acid is written as

Ka =[A-] [H+]/[HA]

Ka = x^2/0.201-x

[A-] = [H+] = x = 3.80 x 10^-5

x is very small, so use 0.201-x as 0.201

Ka = (3.80 x 10^-5)^2/0.201

Ka = 7.18 x 10^-9

2) Ka when pH is 4.84

[H+] = 10^-pH = 10^ -4.84

[H+] = 1.45 x 10^-5

Ka = (1.45 x 10^-5)^2/0.201

Ka = 1.05 x 10^-9

3) Ka when pH is 5.48

[H+] = 10^-pH = 10^ -5.48

[H+] = 3.31 x 10^-6

Ka = (3.31 x 10^-6)^2/0.201

Ka = 5.45 x 10^-11

4) Ka when pH is 10.1

[H+] = 10^-pH = 10^ -10.1

[H+] = 7.94 x 10^-11

Ka = (7.94 x 10^-11)^2/0.201

Ka = 3.14 x 10^-20

5) Ka when pH is 12.11

[H+] = 10^-pH = 10^ -12.11

[H+] = 7.76 x 10^-13

Ka = (7.76 x 10^-13)^2/0.201

Ka = 2.99 x 10^-24

c) Average Ka = ((7.18 x 10^-9) + (1.05 x 10^-9) + (5.45 x 10^-11) + (3.14 x 10^-20) +(2.99 x 10^-24))/5

Average Ka = 3.96 x 10^-15

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