1. A 0.10 M solution of NaCH3COO has a pH of 8.8. Write the equilibrium equation
ID: 891544 • Letter: 1
Question
1. A 0.10 M solution of NaCH3COO has a pH of 8.8. Write the equilibrium equation and then calculate the Kb for the salt, NaCH3COO. Show your work.
2. The pH of a weak acid solution is measured to be 3.50. Calculate its hydrogen ion concentration.
3.The addition of some acids to water generates a considerable amount of heat. What difficulty could this create when trying to determine the Ka value in the lab?
4.A student following the procedure in this lab used a 3.00 M stock solution of an unknown acid to prepare 100 mL of 1.50 M of diluted unknown acid solution. The pH reading of the diluted solution was 3.48. How many mL of unknown 3.00 M stock solution did the student use to prepare the unknown diluted solution?
Explanation / Answer
1)
CH3COO- + H2O ------------------> CH3COOH + OH-
it is a salt of weak acid and strong base. so pH greater than 7
pH = 7 + 1/2 (pKa + log C)
8.8 = 7 + 1/2 (pKa + log 0.1)
pKa = 4.6
Ka = 10^-pKa = 10^-4.6
= 2.51 x 10^-5
Kb = Kw / Ka = 1.0 x 10^-14 / 2.51 x 10^-5
= 3.98 x 10^-10
Kb = 3.98 x 10^-10
2)
pH = 3.50
pH = -log [H+]
3.50 = -log[H+]
[H+] = 3.16 x 10^-4 M
4)
M1 = 3.00 M , V1 = ??
M2 = 1.50 M , V2 = 100 ml
M1 V1 = M2 V2
3 x V1 = 1.5 x 100
V1 = 50 mL
volume of unknowm 3 .00 M stock solution = 50 mL
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