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(.40รท.120)=?
3rd: answer:7pH
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?
2 answers
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.
1 answer