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
Should the pH of a buffer change when the buffer is diluted? Explain fully, using the Henderson-H...
help please!!!!
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...
How do you find the degree of ionization of the solutions when
accounting for the dillution effect of adding 10 mL of water? Also
for the second page, how do you calculate pH with a buffer using
the Henderson-Hasselbalch equation?
Solution 2 - 10.0 mL of 0.1 M HC2H302 +10.0 mL H20 Measured pH_2.42 To determine the [H3O+], account for the dilution effect of adding 10.0 mL of water, and use MV1 = M2V2 Degree of lonization of solution 2...
1. What is the pH of a buffer consisting of 0.30 M CH3COOH & 0.20 M NACH:COO? Ka= 1.8 x 10-5 2. What is the pH of the buffer with 0.10 M NH3 and 0.20 M NH4NO3? Kl = 1.8 x 10-5 3. What is the pH of a buffer formed from combining 10 mL of 0.250 M HCl with 90 mL of 0.150 M NH3? (K = 1.8 x 10-) 4. Calculate the pH of the solution that results...
5. (2pts) To calculate the pH of buffer solution we need to use Henderson-Hasselbalch equation The generic form of this equation is: 6. (8 points) What is the pH of a solution that contains 25 ml of 0.10 M HF and 25 ml 0.1M NaOH solution? (Ka of HF -6.8 x 104).
5. You are asked to prepare an acetate buffer solution at pH 4.6. Use the Henderson-Hasselbalch equation to calculate the ratio of sodium acetate and acetic acid required to prepare a buffer at pH 4.6. 6. If you had 500mL of 0.200M sodium acetate buffer pH4.76, how many mL of 1.00M HCI would you need to add to adjust the pH to 4.60?
buffer is HC2H3O2/NaC2H3O2
difference in PH: big change when added to buffer. small
change when added to water. why?
why such a bug change when you add an acud or base to water
but not to butfer?
1. a) Describe the difference in observed pH changes upon adding a small amount of strong acid or base to 25.0 mL of water vs to 25.0 mL of your buffer. b) Explain why there was a difference. 6.0M HC,H,O, is corrosive. Prevent...
What is the pH value of the following composition of solutions? a) 1 mL 0.10 M HC2H3O2 + 99 mL H2O b) 5 mL 0.10 M HC2H3O2 + 5 mL 0.10 M HCl c) 0.10 M H3PO4 d) 0.10 M NH3 e) 0.10 M NH4NO3 f) 50 mL 0.10 M NH3 + 50 mL 0.10 M NH4NO3 g) 10 mL Solution (h) + 6 mL H2O h) 10 mL Solution (h) + 5 mL H2O + 1 mL 0.10 M...
1 . If a buffer solution is 0.260 M in a weak acid (?a=8.3×10−5)and 0.480 M in its conjugate base, what is the pH? pH= 2. If a buffer solution is 0.200 M in a weak base (?b=5.0×10−5) and 0.530 M in its conjugate acid, what is the ph 3. Phosphoric acid is a triprotic acid (?a1=6.9×10−3, ?a2=6.2×10−8 , and ?a3=4.8×10−13 To find the pH of a buffer composed of H2PO4 - (aq) ) and HPO4 2− (aq) , which...
1) Calculate the pH of the 1L buffer composed of 500 mL of 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of HCl is added (Ka HC2H3O2 = 1.75 x 10-5). Report your answer to the hundredths place. 2) Calculate the pH of the 1L buffer composed of 500 mL 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of NaOH is added (Ka HC2H3O2 = 1.75...
1)
A buffer can be prepared by mixing two solutions. Determine if
each of the following mixtures will result in a buffer solution or
not.
1) Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M NaOH
[
Select ]
["No, it will not result in
a buffer solution.", "Yes, it will result in a buffer solution."]
2) Mixing...