Ka for benzoic acid, C6H5COOH, 6.5x10^-5. Calculate the pH of solution after addition of 10.0, 20.0, 30.0, and 40.0 mL of 0.10 M NaOH to 40.0 mL of 0.10 M Benzoic acid.
PLEASE CHECK MY ANSWER!!!!!
My answer is:
Moles acid = 0.040 L x 0.10 M = 0.0040
Moles base = 0.010 L x 0.10 M = 0.0010
C6H5COOH + OH- >> C6H5COO- + H2O
moles acid = 0.0040 -0.0010 = 0.0030
moles salt = 0.0010
total volume = 10 + 40 = 50 mL = 0.050 L
concentration acid = 0.0030 / 0.050 =0.060 M
concentration salt = 0.0010 / 0.050 =0.020 M
pKa = 4.19
pH = 4.19 + log 0.020 / 0.060 =3.71
work in the same way I get answers for the other questions.
Looks good to me. Thanks for showing your work. I was about to pass it by until I saw you wanted an answer check.
thank you very much, DrBob222!
Your calculations are correct for the first question.
For the second question:
Moles acid = 0.040 L x 0.10 M = 0.0040
Moles base = 0.020 L x 0.10 M = 0.0020
C6H5COOH + OH- >> C6H5COO- + H2O
moles acid = 0.0040 - 0.0020 = 0.0020
moles salt = 0.0020
total volume = 20 + 40 = 60 mL = 0.060 L
concentration acid = 0.0020 / 0.060 = 0.033 M
concentration salt = 0.0020 / 0.060 = 0.033 M
pKa = 4.19
pH = 4.19 + log (0.033 / 0.033) = 4.19 + log (1) = 4.19
Your calculations for the concentration of acid and salt are incorrect. Both should be 0.0020/0.060 = 0.033 M. Therefore, the pH would be the same as the pKa which is 4.19.
Please make sure to double-check your calculations for the other questions as well.
To calculate the pH of the solution after the addition of NaOH, you first need to determine the concentration of the benzoic acid (C6H5COOH) and the benzoate ion (C6H5COO-) after the neutralization reaction.
1. Moles of benzoic acid:
Volume of benzoic acid = 40.0 mL = 0.040 L
Concentration of benzoic acid = 0.10 M
Moles of benzoic acid = Volume x Concentration = 0.040 L x 0.10 M = 0.0040 mol
2. Moles of NaOH:
Volume of NaOH added = 10.0 mL = 0.010 L
Concentration of NaOH = 0.10 M
Moles of NaOH = Volume x Concentration = 0.010 L x 0.10 M = 0.0010 mol
3. Since benzoic acid and NaOH react in a 1:1 ratio, the moles of benzoic acid that have reacted can be calculated by subtracting the moles of NaOH from the moles of benzoic acid:
Moles of benzoic acid remaining = 0.0040 mol - 0.0010 mol = 0.0030 mol
4. Moles of benzoate ion:
Since the reaction is 1:1, the moles of benzoate ion formed is equal to the moles of NaOH added:
Moles of benzoate ion = 0.0010 mol
5. Total volume of the solution:
Volume of benzoic acid + Volume of NaOH added = 40.0 mL + 10.0 mL = 50.0 mL = 0.050 L
6. Concentration of benzoic acid and benzoate ion:
Concentration of benzoic acid = Moles of benzoic acid / Total volume = 0.0030 mol / 0.050 L = 0.060 M
Concentration of benzoate ion = Moles of benzoate ion / Total volume = 0.0010 mol / 0.050 L = 0.020 M
7. pKa value for benzoic acid = 4.19
8. pH calculation:
pH = pKa + log10 (concentration of benzoate ion / concentration of benzoic acid)
pH = 4.19 + log10 (0.020 M / 0.060 M)
pH = 4.19 + log10 (1/3)
pH = 4.19 - log10 (3)
pH ≈ 4.19 - 0.48
pH ≈ 3.71
Therefore, the pH of the solution after adding 10.0 mL of 0.10 M NaOH to 40.0 mL of 0.10 M benzoic acid is approximately 3.71.
You can follow the same steps to calculate the pH after adding 20.0 mL, 30.0 mL, and 40.0 mL of 0.10 M NaOH to the same 40.0 mL of 0.10 M benzoic acid. Remember to adjust the volumes and moles accordingly in each calculation.