Practice Problems Week 6 p H and Titrations - With Answer Key PDF

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Study Session 2 – Lab Titrations and pH Calculations 1. According to the Strong Acid (HCl)/ Strong Base (NaOH) titration in this week’s lab, using the indicator method, what was the concentration of the HCl from the data?

2. According to the Weak Acid (HC2H3O2)/ Strong Base (NaOH) titration using an indicator method, what was the concentration of the Acetic acid from the data?

3. According to the Lab data, what was the concentration of NaOH prepared for the strong acid and weak acid titrations?

4. According to the Lab data, what was the concentration of NaOH prepared for the Buffer portion of the lab?

5. According to the Lab data, what was the concentration of Acetic Acid prepared for the Buffer portion of the lab?

6. According to the Lab data, what was the concentration of Sodium Acetate prepared for the Buffer portion of the lab?

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7. Imagine a new lab scenario. You took a clean and dry crucible and measured it to be 25.125 grams. To the crucible, you added 50.0 mL of NaCl solution and placed the solution on a hot plate to evaporate all the water away. You then measured the crucible with solid sodium chloride in it to find that the mass was 28.778 grams. What was the concentration of the NaCl solution?

Definitions (relationship between pX and concentration of [X]

Relationships between H+ and OH-

pH = -log[H+]

[H+][OH-] = 1 x 10-14

[H+] = 10-pH

pOH = -log[OH-] [OH-] = 10-pOH

pH + pOH = 14

8. Find the pH of an acidic solution with [H+] = 1.3 x 10-4 M

9. Find the pH of a basic solution with [H+] = 2.75 x 10-12 M

10. Find the pH of a basic solution if pOH = 3.77

11. Find the pOH of an basic solution of the [OH-] = 0.0125 M

12. Find the pH of an basic solution of the [OH-] = 0.0125 M

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13. What is the pH of 50.0 mL of 0.01 M HCl strong acid solution?

14. What is the pH of 25.0 mL of 0.01 M H2SO4 strong acid solution?

15. What is pH of 2.2 x 10-4 M Ca(OH)2 strong base solution?

Study Session 3 – ICE Tables of weak acids and weak bases Example: You calculated in problem #13 that you’d expect a pH = 2 for a 0.01 M strong acid solution. But the calculation is not as straightforward for 0.01 M weak acid solution since a weak acid only partially ionizes, therefore we can’t say concentration of weak acid is equal to concentration of H+. Acid equilibrium (Ka values) and ICE tables are now required to determine the pH of a weak acids and weak bases. Find the pH of 0.01 M solution of acetic acid (Ka = 1.8 x 10-5) C2H3O2 -

HC2H3O2

H+

0.01

-

-

-x

+x

+x

0.01 – x

x

x

[𝑯+][𝑪𝟐𝑯𝟑𝑶𝟐−]

Ka = (

[𝑯𝑪𝟐𝑯𝟑𝑶𝟐]

1.8 x 10-5 = (

[𝒙][𝒙]

)

[𝟎.𝟎𝟏−𝒙]

)

Approximate (denominator – check that concentration of x is less than 5% of 0.01 to confirm that the approximation was a valid thing to do) [𝒙][𝒙] ) 𝟎.𝟎𝟏]

1.8 x 10-5 = ([

Solve for x = 4.24 x 10-4 (x represents [H+] so pH = -log (4.24 x 10-4) = 3.373 3

16. Nitric acid is a strong acid while nitrous acid is a weak acid (Ka = 4.6 x 10-4). Determine the pH of 0.200 M HNO3 (aq)

17. Determine the pH of 0.200 M HNO2 (aq)

18. What is the pH of water if the temperature is increased to 30.0∘C, where the Kw = 1.47 × 10-14

19. (a) What is the pH for an aqueous solution of NH3 that has a hydroxide ion concentration of 2.25 × 10-2 M at equilibrium. The Kb = 1.76 x 10-5 NH3 + H2O → NH4 + OH-

(b) What is the pH if the concentration of ammonia is 0.100 M NH3

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Study Session 4 – Buffers Buffer Solutions Contain both weak acids and their conjugate bases. You can use the Henderson-Hasselbalch equation to determine the pH of a buffer solution (under the 5% “x is small” approximation)

Henderson-Hasselbalch equation [𝐜 𝐛𝐚𝐬𝐞]

pH = 𝐩𝐊𝐚 + 𝐥𝐨𝐠 ([𝐰 𝐚𝐜𝐢𝐝]) Example: Calculate the pH of a buffer solution that is 0.050 M Benzoic acid (HC7H5O2) and 0.150 M sodium benzoate (NaC7H5O2). The Ka of benzoic acid is 6.5 x 10-5 Class Example pKa = -log(Ka) = -log(6.5 x 10-5) = 4.187 [𝒄 𝒃𝒂𝒔𝒆]

pH = 𝒑𝑲𝒂 + 𝒍𝒐𝒈 ([𝒘 𝒂𝒄𝒊𝒅]) pH = 4.187 + 𝑙𝑜𝑔 (

[0.150]

[0.050]

)= 4.66

Check if approximation is valid HA

H+

A0.150

0.05

0.150

0.05 – x

+x

[𝒙][𝟎.𝟏𝟓𝟎+𝒙]

Ka = (

[𝟎.𝟎𝟓−𝒙]

0.150 + x

)

Approximate [𝒙][𝟎.𝟏𝟓𝟎]

6.5 x 10-5 = (

[𝟎.𝟎𝟓]

)

Solve for x = 2.17 x 10-5 pH = -log(2.17 x 10-5) =4.66 2.17 x 10-5 / 0.050 = valid!

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20. Preparing a buffer with a target pH Make a buffer with bicarbonate ion serving as the weak acid (conjugate base is carbonate ion). You are given a restriction that the concentration of weak acid is 0.10 M. What concentration of conjugate base should you use to make a buffer with pH = 9.9? The Ka of HCO3- = 4.677 x 10-11 (The source of the acid will come from NaHCO3. The source of the conjugate base will come from Na2CO3)

HCO3- → H+ + CO32-

21. Preparing a Buffer with a target pH Make a buffer with acetic acid serving as the weak acid (conjugate base sodium acetate). You are given a restriction that the concentration of conjugate base is 0.20 M. What concentration of weak acid should you use to make a buffer with pH = 4.4? The pKa of acetic acid is 4.74

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Study Session 5 – Lab 22. Calculate the strong acid concentration from the pH titration curve method (Use the volume of base added from the double derivative graph) and compare it to the concentration you determined from the titration where you used an indicator.

23. Consider the weak acid titration where you made a titration curve. From the equivalence point, find the ½ equivalence point which will guide you to where pH = pKa. From the pH value you pull off of the graph at the midpoint, find the Ka of the acetic acid.

24. In the lab video, you saw how by adding strong base to a weak acid, the pH increases. The next series of questions walks you through a few problems to theoretically calculate the pH at different stages of a titration. (this is not the buffer solution titration, this is the acetic acid/NaOH titration) Consider the first step. Calculate the pH of the acetic acid in the beaker before adding any base at all. The pH meter gave a reading which is the “experimental” pH, but you are being asked to calculate the theoretical pH. (hint, use an ice table)(compare it to the pH meter reading)

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25. Now consider the pH at the equivalence point. To understand why it is above pH= 7, you need to consider that all the OH- has matched up with all the H+, but what is left in solution is the conjugate base of the acetic acid (the acetate anion). The acetate ion reacts with water in the same equilibrium (but reversed) as acetic acid ionization. The strategy is to set up a Kb expression, where x will be equal to [OH-]. Then solve for the pH as you did in Study Session 2 using information about [OH-]. Recall that Kw = KaKb. The acetic acid Ka = 1.8 x 10 -5 A+ H2O → HA + OH(concentration – x) +x +x The concentration of A- comes from the following formula concentration A- = (Volume of NaOH at equivalence point) (concentration of NaOH) Volume of everything in beaker

26. Now consider the pH for each data point between the first point and the equivalence point. This is the buffer region of a titration curve and you can approximate the pH using the Henderson-Hasselbalch equation. What is the theoretical pH after adding 3.20 mL of NaOH to the acid? Starting from here :

[𝒄 𝒃𝒂𝒔𝒆]

pH = 𝒑𝑲𝒂 + 𝒍𝒐𝒈 ([𝒘 𝒂𝒄𝒊𝒅])

Convert the [c base] and [w acid] into “moles” of base, and “moles” of acid that are in the beaker, Hint: “moles of c. base” = 0.00. This would be a non-zero number if you prepared an actual buffer solution Then solve for pH :

pH = 𝒑𝑲𝒂 + 𝒍𝒐𝒈 (

𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒄 𝒃𝒂𝒔𝒆 + 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑵𝒂𝑶𝑯 𝒂𝒅𝒅𝒆𝒅 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒘 𝒂𝒄𝒊𝒅−𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑵𝒂𝑶𝑯 𝒂𝒅𝒅𝒆𝒅

)

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27. Finally, determine the theoretical pH beyond the equivalence point when volume of base added = 10.46 mL NaOH added (use 10.32 mL as equivalence point for this problem). Here, the strategy will be that you consider how many OH’s added beyond the equivalence point to determine the number of free OH- in solution that have nothing to react other than the equilibrium with water. a. Find the volume of NaOH beyond the equivalence point, and multiply it by the NaOH concentration to determine the number of free moles of OH-

b. Next find the concentration of OH- by putting (moles of OH-) over the total volume in the beaker at this point.

c. Calculate the pH from the [OH-]

28. Using the Henderson- Hasselbalch equation, what was the theoretical pH of the buffer solution made in the lab (Questions 5 and 6 of this packet may help)

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Answer Key: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.

24. 25. 26. 27. 28.

0.198 M HCl 0.103 M HC2H3O2 Given to you, no calculation needed = 0.148 M 0.200 M NaOH 0.110 M HC2H3O2 0.200 M NaC2H3O2 1.25 M NaCl solution 3.89 11.561 10.23 1.903 12.097 2.000 1.7 (2 H’s per H2SO4 ) 10.64 0.699 2.02 6.92 a.12.35 (basic because of OH- ions that form) b. 11.123 0.037 M 0.438 M 13.36 mL from graph gives the same concentration = 0.198 M From my graph the ½ equivalence point is 5.16 mL which corresponds to a pH = 4.71 and gives a Ka = 1.95 x 10-5. Depending on how much you zoomed in on x and y-axis, your value may vary slightly from mine. pH = 2.87 between pH = 8.76 and 8.77 depending on whether you used 10.32 mL or 10.40 mL of base added pH = 4.40 depending on which volume of base you used from either the indicator titration or titration curve method, the number of free moles of OH- ions range from 8.88 x 10-6 to 2.07 x 10-5 pH = 5.00

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