Asked by KO
2. Determine the pH if 0.02 mol of HCl is added to 1.0 L of the buffer described in question #1
question1: 1. How many grams of NaHCO3 should be added to one liter of 0.100 M H2CO3 (Ka = 4.2 x 10-7) to prepare a buffer with pH = 7.00?
question1: 1. How many grams of NaHCO3 should be added to one liter of 0.100 M H2CO3 (Ka = 4.2 x 10-7) to prepare a buffer with pH = 7.00?
Answers
Answered by
Dr. Neutron
1st: equation: H2CO3+HCl<-->H2CO3+Cl
2ndICE For Initial: H2CO3=.42, HCl=.02,H2CO3=..100from?1
For Change:H2CO3=-.02, HCl=-.02,H2CO3=.+.02
For Equilium:H2CO3=.40, HCl=0,H2CO3=.120
2nd:Use Buffer equation: pH= -log(4.2x10-7)+log(.40÷.120)=?
3rd: answer:7pH
2ndICE For Initial: H2CO3=.42, HCl=.02,H2CO3=..100from?1
For Change:H2CO3=-.02, HCl=-.02,H2CO3=.+.02
For Equilium:H2CO3=.40, HCl=0,H2CO3=.120
2nd:Use Buffer equation: pH= -log(4.2x10-7)+log(.40÷.120)=?
3rd: answer:7pH
Answered by
Doc
Using the Henderson-Hasselblach equation => Salt to Acid Ratio =[NaHCO3]/H2CO3] = 0.238/1. Thus, for a .1M H2CO3 the buffer would be 0.0238M in NaHCO3. => grams of NaHCO3 needed = 0.0238 mole(84 g/mole) = 20 grams.
Adding 0.02mole HCl to the buffer solution => [HCl]added = 0.02M and would shift the equilibrium left.
H2CO3 <=> H^+ + HCO3^-
0.100M - 0.0238M
+0.020M - -0.0200M
0.120M [H^+] 0.0038M
Solve Ka expression for [H^+] = 1.33E-5M => pH = -log(1.33E-5) = 4.88 after adding 0.02 mole HCl into the 1.0 Liter of pH = 7 Buffer.
Adding 0.02mole HCl to the buffer solution => [HCl]added = 0.02M and would shift the equilibrium left.
H2CO3 <=> H^+ + HCO3^-
0.100M - 0.0238M
+0.020M - -0.0200M
0.120M [H^+] 0.0038M
Solve Ka expression for [H^+] = 1.33E-5M => pH = -log(1.33E-5) = 4.88 after adding 0.02 mole HCl into the 1.0 Liter of pH = 7 Buffer.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.