Date REPORT SHEET I EXPERIMENT Titration Curves of 26 Poly protic Acids B. Stand

Question

Date REPORT SHEET I EXPERIMENT Titration Curves of 26 Poly protic Acids B. Standardization of Sodium Hydroxide (NaOH) Solution Trial 2 Mass of bottle +KHP Mass of bottle Mass of KHP used Final buret reading tial buret reading mL of NaOH used Molarity of NaOH Average molarity (show calculations) Standard deviation (show calculations) C. Determination of the Acid Dissociation Constants and the Molarity of the Unknown Acid Trial I Trial 2 Volume of unknown acid Molarity of NaOH from above 3yasM ml of NaOH at eq- point i ml., of NaOH at eq. point 2 Molarity of unknown acid Copyright c 2015 Pearson Education, Inc.

Explanation / Answer

Standard Deviation:

Actually In the above we have molarity of NAOH is 3475

and the averge molarity is 3475.65

then standard deviation is

SD= {([value-mean]2)/(number of trials -1)

(3475-3475.65)^2 /1

0.4225.

b) from the above the moles is 3475

and volume is 25L

then avg molarity is 3475/25=139

then standard deviation is 33.34

c) find pka

DATA:
MM = 71.8-g/mol
W = 7.20-g
V = 1-L
n = 7.2/71.8=0.1003
Mi=0.1003-mol/1-L=0.1003-M
pH = 2.110
[H+] = 10^-pH = 10^-2.110 = 0.0078

Considering all is in the same volume, I mean 1-L:

...........HA... =====> H+...+...A-
i.......0.1003
e......0.0078
f.......0.0925...............0.0078......

In the equilibrium:

[H+] = 0.0078
.[A-] = 0.0078
[HA] = 0.0925

From here we can say:

Ka = [H+][A-]/[HA]
-LogKa = -Log([H+][A-]/[HA])
......pKa = -Log(0.0078x0.0078/(0.1003-0.0078))
......pKa = 3.19

To calculate the pH of an aqueous solution you need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is then calculated using the expression:

pH = - log [H3O+].

Example: Find the pH of a 0.0025 M HCl solution. The HCl is a strong acid and is 100% ionized in water. The hydronium ion concentration is 0.0025 M. Thus:

pH = - log (0.0025) = - ( - 2.60) = 2.60

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Calculating the Hydronium Ion Concentration from pH

The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH.

[H3O+] = 10-pH or [H3O+] = antilog (- pH)

Example: What is the hydronium ion concentration in a solution that has a pH of 8.34?

8.34 = - log [H3O+]
- 8.34 = log [H3O+]
[H3O+] = 10-8.34 = 4.57 x 10-9 M

On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34).

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Calculating pOH

To calculate the pOH of a solution you need to know the concentration of the hydroxide ion in moles per liter (molarity). The pOH is then calculated using the expression:

pOH = - log [OH-]

Example:  What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10-5 M?

pOH = - log [4.82 x 10-5] = - ( - 4.32) = 4.32

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Calculating the Hydroxide Ion Concentration from pOH

The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH.

[OH-] = 10-pOH   or   [OH-] = antilog ( - pOH)

Example:  What is the hydroxide ion concentration in a solution that has a pOH of 5.70?

5.70 = - log [OH-]
-5.70 = log[OH-]
[OH-] = 10-5.70 = 2.00 x 10-6 M

On a calculator calculate 10-5.70, or "inverse" log (- 5.70).

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Relationship Between pH and pOH

The pH and pOH of a water solution at 25oC are related by the following equation.

pH + pOH = 14

Write the dissociation equation for the acid:

HA H+ + A¯

2) Write the equilibrium expression:

Ka = ( [H+] [A¯] ) / [HA]

3) Our task now is to determine the three concentrations on the right-hand side of the equilibrium expression since the Ka is our unknown.

a) We will use the pH to calculate the [H+]. We know pH = -log [H+], therefore [H+] = 10¯pH

[H+] = 10¯3.26 = 5.4954 x 10¯4 M

I've kept a couple guard digits; I'll round off the final answer to the proper number of significant figures.

b) From the dissociation equation, we know there is a 1:1 molar ratio between [H+] and [A¯]. Therefore:

[A¯] = 5.4954 x 10¯4 M

c) the final value, [HA] is given in the problem. In the example being discussed, 0.120 M is the value we want. Some teachers will use 0.120, while others would say to subtract the 5.4954 x 10¯4value from 0.120 first. Let's do both.c1) Ka = [(5.4954 x 10¯4) (5.4954 x 10¯4)] / 0.120

Ka = 2.52 x 10¯6

c2) Ka = [(5.4954 x 10¯4) (5.4954 x 10¯4)] / (0.120 minus 5.4954 x 10¯4)

Ka = 2.53 x 10¯

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