Question

please help me answer those 3 questions!!

Should the pH of a buffer change when the buffer is diluted? Explain fully, using the Henderson-Hasselbalch equation as wel as your results from Solutions 10 and 11 c. pH calculation for Solution 13 pH calculation for Solution 14 Sol'n Composition 15 mL CH3COOH 2 10 mL CH3COOH+10 mL H20 3 1 mL CH3COOH + H20 to make 100 mlL 4 10 mL CH3COOH +10 mL HCI 5 H3PO4 6 7 8 10 mL CH3COOH +5 mL NaOH 9 10 mL NH4NO3 +5 mL NaOH NH3 7 NH4NO 10 11 12 50 mL NH3 +50 mL NH4NO3 10 mL Solution 10+ 6 mL H20 10 mL Solution 10 +5 mL H2O+ 1 mL HC 13 10 mL Solution 10+ 6 mL HCI 14 10 mL Solution 10 +5 mL H2O+1 mL NaOH Set 1 Set 2 Set 3 Solution No PH Solution No. PH Solution No. pH . 2. 85 3.01 S.4S 1.76 10 t04 6 10. 3 5.457 6. 40 9 1.17 12 .6C 4 13 14

Answer 1

pH calculation for solution 13.

Solution 10: The no. of millimoles of NH₃ = 50 mL * 0.1 mmol/mL = 5 mmol and the no. of millimoles of NH₄NO₃ = 50 mL * 0.1 mmol/mL = 5 mmol

The volume of the total solution = 50+50 = 100 mL

Now, the no. of millimoles of NH₃ = 5/10 = 0.5 mmol and the no. of millimoles of NH₄NO₃ = 5/10 = 0.5 mmol

The molarity of HCl is not given, let's consider 0.01 M.

The no. of millimoles of HCl = 6 mL * 0.01 mmol/mL = 0.06 mmol

The resulting millimoles of NH₃ = 0.5-0.06 = 0.44 mmol and the resulting millimolesof NH₄NO₃ = 0.5+0.06 = 0.56

Now, according to the Henderson-Hasselbulch equation:

pH = pKa + Log([NH₃]/[NH₄NO₃])

Here, pKa of ammonia = 9.26

Therefore, pH = 9.26 + Log(0.44/0.56) = 9.16

pH calculation for solution 14.

Solution 10: The no. of millimoles of NH₃ = 50 mL * 0.1 mmol/mL = 5 mmol and the no. of millimoles of NH₄NO₃ = 50 mL * 0.1 mmol/mL = 5 mmol

The volume of the total solution = 50+50 = 100 mL

Now, the no. of millimoles of NH₃ = 5/10 = 0.5 mmol and the no. of millimoles of NH₄NO₃ = 5/10 = 0.5 mmol

The molarity of NaOH is not given, let's consider 0.01 M.

The no. of millimoles of NaOH = 1 mL * 0.01 mmol/mL = 0.01 mmol

The resulting millimoles of NH₃ = 0.5+0.01 = 0.51 mmol and the resulting millimolesof NH₄NO₃ = 0.5-0.01 = 0.49

ow, according to the Henderson-Hasselbulch equation:

pH = pKa + Log([NH₃]/[NH₄NO₃])

Here, pKa of ammonia = 9.26

Therefore, pH = 9.26 + Log(0.51/0.49) = 9.28

Question 3.

No, the pH of a buffer should not change when the buffer is diluted.

Explanation: By the dilution, the no. of millimoles will be the same.

Solution 10. pH = 9.26 + Log(5/5) = 9.26

Solution 11. pH = 9.26 + Log(0.5/0.5) = 9.26