Question

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3. Compare Solutions 2 and 4. How does the common-ion effect influence the pH of Solution 2? Explain fully, and calculate the expected pH of these solutions pH calculation for Solution 2 a. pH calculation for Solution 4 Your explanation 4. What would be the pH of solutions 8 and 9? Give a relevant chemical equation and related PH calculation. Compare the calculated value properties of a buffer? eeio Cocal eqatind eld ?5. a. How do Solutions 11 14 show the and experimental value. a. Solution 8 b. Calculate the expected pH for each of these solutions pH calculation for Solution 11 b. Solution 9 pH calculation for Solution 12 pH calculation for Solution 13 Should the pH of a buffer change when the buffer is diluted? Explain fully, using the Henderson-Hasselbalch equation as well as your results from Solutions 10 and 11. c. pH calculation for Solution 14 pH data Set 1 Set 2 Set 3 pH 2. 85 3.01 S.4S 1.76 Solution No. pH Solution No. pH 10 2.04 2 40 12 .CC SI 6. 9o 4 8 13 9 .17141.03 Table 8.1 Solutions to be investigated. Concentration of all solutions is 0.10 M Sol'n Composition 1 15 mL CH3COOH 2 10 mL CH3COOH + 10 mL H2O 3 1 mL CH3COOH + H2O to make 100 mL 4 10 mL CH3COOH +10 mL HCI H3PO4 6 NH3 7 NH4NO3 8 10 mL CH3COOH +5 mL NaOH 10 11 12 9 10 mL NH4NO3 5 mL NaOH 50 mL NH3 +50 mL NH4NO3 10 mL Solution 10 +6 mL H2O 10 mL Solution 10 +5 mL H20+1 mL HCI 13 10 mL Solution 10 +6 mL HC 14 10 mL Solution 10 +5 mL H20 1 mL NaOH

Answer 1

#3: pH calculation for solution-2:

Total volume = 10 mL + 10 mL = 20 mL * (1L / 1000 mL) = 0.020 L

Since equal volume of water and CH3COOH are added, the concentration will be halved.

=> [CH₃COOH] = 0.10 / 2 = 0.050 M

---CH₃COOH <-----> CH₃​​​​​​​COO^-(aq) + H⁺

I: 0.050 ------------------ 0 ------------------ 0

C: - X -------------------- +X ---------------- +X

E: (0.050 - X), ---------- X ------------------- X

Ka = 1.8*10^-5 = X.X / (0.050 - X)

Since X << 0.050

=> 1.8*10^-5 = X² / 0.050

=> X² = 0.050*1.8*10^-5

=> X = [H⁺]= 9.49*10^-4

=> pH = - log[H⁺] = - log(9.49*10^-4 ) = 3.02 (Answer)

pH calculation for solution-4:

Total volume = 10 mL + 10 mL = 20 mL * (1L / 1000 mL) = 0.020 L

Since equal volume of HCl and HCl are added, the concentration of both will be halved.

=> [CH₃COOH] = 0.10 / 2 = 0.050 M

Since HCl is a strong acid it is completely dissolved. Hence [HCl] = initial [H⁺] = 0.10 / 2 = 0.050 M

---CH₃​​​​​​​COOH <-----> CH₃​​​​​​​COO^-(aq) + H⁺

I: 0.050 ------------------ 0 ------------------ 0.050

C: - X -------------------- +X ---------------- +X

E: (0.050 - X), ---------- X ------------------- X+0.050

Ka = 1.8*10^-5 = X.(X+0.050) / (0.050 - X)

Since X << 0.050, we can neglect X

=> 1.8*10^-5 = X*0.050 / 0.050

=> X = 1.8*10^-5

[H⁺] = X+0.050 = 1.8*10^-5 + 0.050 $\approx$ 0.050

=> pH = - log[H⁺] = - log(0.050) = 1.30 (Answer)

Explanation: Due to the common ion H⁺, the dissociation CH3COOH in very low in solution-4 in comparison to solution-2.   

In solution-2, X = 9.49*10^-4 , where as in solution-4, X = 1.8*10^-5