Question

can someone please me with checking our work for the chart and helping me with the missing ones. Also with question 4 and 5? Thank You

Concentration of standardized NaOH Volume of acid used NaOH volume added before largest pH increase NaOH volume added after largest pH increase pH before the largest pll increase pH after the largest pH increase I HC HC,H,O, 0.100 M 0.100 M 10.0 ml. 10 ml 10.2 ml. 8.6 ml. 112 ml 100 ml 5.495 .98 10.90 10.68 Volume of NaOH added at equivalence point 10.7 ml. 9.5 mL pH at equivalent point 10.38 10.02 Moles NaOH mol mol Moles of acid mol mol Concentration of acid Vml/2 pka

Answer 1

The moles of NaOH are calculated:

n NaOH = M * V = 0.1 M * 0.0107 L = 0.00107 mol

Trial 2: 0.00095 mol

The moles of acid are equal to the moles of NaOH.

The acid concentration is calculated:

[Acid] = n / V = ​​0.00107 / 0.01 = 0.107 M

Trial 2: 0.095 M

The V / 2 = 9.5 / 2 = 4.75 mL is read

The pH = 4.66 = pKa is read

5) The pKa value of the literature is 4.75, the experimental value is quite close.

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.