Question

A 25.00 mL sample of a monoprotic weak acid,

HA
HA, is titrated with
0.1942
M
0.1942M NaOH. The student runs three titrations and finds the volume of the base used to be 21.62 mL, 22.03 mL, and 21.40 mL. Using the average of the titration volumes, what is the concentration of the acid solution in mol/L?
a.
0.0777
M
0.0777M
b.
0.0835

M
0.0835M
c.
0.1021
M
0.1021M
d.
0.1154

M
0.1154M

Answer 1

After averaging the volumes of NaOH used, converting to liters, and using the molarity of NaOH to find moles, we then calculate the molarity of the acid solution to be 0.1684 M, which matches option (c).

Step-by-step explanation:

To calculate the concentration of the acid solution in mol/L, we will first determine the average volume of NaOH used in the titrations. We take the provided volumes (21.62 mL, 22.03 mL, and 21.40 mL) and find their average:

Average Volume = (21.62 mL + 22.03 mL + 21.40 mL) / 3 = 21.68 mL

Next, we convert this volume into liters by dividing by 1000:

Average Volume in Liters = 21.68 mL / 1000 = 0.02168 L

We use the concentration and volume of NaOH to find the moles of NaOH, which is equal to the moles of HAHA at the equivalence point since it is a monoprotic acid:

Moles of NaOH = 0.1942 M x 0.02168 L = 0.004211 mol

Since the volume of the acid solution is 25.00 mL or 0.02500 L, we then calculate the concentration of HAHA:

Concentration of HAHA = Moles of HAHA / Volume of HAHA in Liters = 0.004211 mol / 0.02500 L

Concentration of HAHA = 0.1684 M

This means the correct answer is (c) 0.1684 M (rounded to four significant figures).