1. The pH of a solution made by the addition of 45mL of 0.100 M NaOH and 50 mL o

Question

1. The pH of a solution made by the addition of 45mL of 0.100 M NaOH and 50 mL of a 0.100M weak acid is determined to be 4.8. What is the Ka of the acid?'

(my answer for this is pKa=3.85) Please help on last 3 questions!!!

2. Given the Ka values of some weak acids, identify the acid being titrated

3. Has equivelance point been reached in above solution? Explain

4. Would you expect pH of the solution to be greater than 7 or less than 7 at the equivalence point? explain

ACID Ka Citric 7.4e-4 Hydrofluoric 6.8e-4 nitrous 4.5e-4 formic 1.8e-4 lactic 1.4e-4 benzoic 6.3e-5 acetic 1.8e-5 hydrocyanic 4.9e-10

Explanation / Answer

1.HA + NaOH ===? NaA + H2O
0.045 liters x 0.10 moles NaOH / liter = 0.0045 moles NaOH
0.0045 moles of NaOH will react with 0.0045 moles of HA and produce 0.0045 moles of A-

moles of weak acid: 0.050 liters x 0.10 moles HA / liter = 0.0050 moles HA
moles of HA remaining after NaOH = 0.0050

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