Question

Titration of weak acid with strong base pre-lab

Answers can only be entered/graded between 2019-10-24 and 2019-10-29 A student peforms a titration, titrating 25.00 mL of a weak monoprotic acid, HA, with a 1.12 M solution of NaOH. They collect data, plot a titration curve and determine the values given in the below table. ml NaOH PH added Half-way Point 17.97 3.66 Equivalence point 35.938.65 How many moles of NaOH have been added at the equivalence point? r mol correct 0.6/1 (sfig err) What is the total volume of the solution at the equivalence point? mL incorrect 0/1

Answer 1

The moles of NaOH are calculated:

n NaOH = M * V = 1.12 M * 0.03593 L = 0.040 mol

You have the total volume of solution:

Vt = 25 + 35.93 = 60.93 mL

The concentration of A- is calculated:

[A-] = n / V = ​​0.04 / 0.06093 = 0.656 M

[H3O +] is calculated:

[H3O +] = 10 ^ -pH = 10 ^ -8.65 = 2.2x10 ^ -9 M

[HA] = 1.12 M * 35.93 mL / 25 mL = 1.61

Ka is calculated:

Kb = [H3O +] * [A-] / [HA]

Kb = 2.2x10 ^ -9 * 0.656 / 1.61 = 9x10 ^ -10

Ka = 10 ^ -14 / 9x10 ^ -10 = 1.1x10 ^ -4

Ka is calculated by the midpoint of equivalence:

Ka = 10 ^ -pKa = 10 ^ -3.66 = 2.2x10 ^ -4

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