A buffer solution is made that is 0.445 M in CH 3 COOH and 0.445 M in CH 3 COO .
ID: 1004211 • Letter: A
Question
A buffer solution is made that is 0.445 M in CH3COOH and 0.445 M in CH3COO .
(1) If Ka for CH3COOH is 1.80×10-5, what is the pH of the buffer solution?
(2) Write the net ionic equation for the reaction that occurs when 0.117 mol HCl is added to 1.00 L of the buffer solution.
Use H3O+instead of H+.
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A buffer solution is made that is 0.496 M in H2S and 0.496 M in KHS.
(1) If Ka for H2S is 1.00×10-7, what is the pH of the buffer solution?
(2) Write the net ionic equation for the reaction that occurs when 0.114 mol HCl is added to 1.00 L of the buffer solution.
Use H3O+ instead of H+.
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A buffer solution is made that is 0.349 M in H2C2O4 and 0.349 M in NaHC2O4.
(1) If Ka for H2C2O4 is 5.90×10-2, what is the pH of the buffer solution?
(2) Write the net ionic equation for the reaction that occurs when 0.098 mol KOH is added to 1.00 L of the buffer solution.
Explanation / Answer
1) CH3COOH+H2O----->CH3COO-+H3O
0.445M 0.445 M
Ka=1.8*10-5={CH3O2-}[H3O+]/[CH3CO2H]
By handerson eqation
pH=pKa+Log[A-]/[HA]
=4.74+Log 0.445/0.445
=4.74
2)CH3COO-+H+--->CH3COOH
before rxn 0.445 M+ 0.117--->0.445M
deta -0.117 -0.117---->+0.117
After RXn 0.328M 0---->0.562M
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H2S +KOH--->KHS+H2O
0.496 0.496M
Ka=1*10-7={H2S] +(KOH)/[KHS]
By handerson eqation
pH=pKa+Log[A-]/[HA]
=6+Log 0.4690.469
=4.74
KHS-+H+--->H2S+KCl
before rxn 0.496 M+ 0.114--->0.496M
deta -0.114 -0.114---->+0.114
After RXn 0.382M 0---->0.61M
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H2CO3+NaOH---->NAHC2O4+H2o
0.349 0.349M
Ka=5.9*10-2=NaHC2O4)/(H2CO3)(OH-)
By handerson eqation
pH=pKa+Log[A-]/[HA]
=1.77+log(0.349/0.349)
=1.77=pH
NAHC2O4+H+--->H2CO3+H20
before rxn 0.349 M+ 0.098--->0.349M
deta -0.098 -0.098---->+0.098
After RXn 0.251M 0---->0.447M
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