Question

A titration setup was used to determine the unknown molar concentration of a solution of NaOH. A1.2 M HCl solution was used as the

titration standard. The following data were collected.
Trial 1
Amount of HCI
Standard Used 10.0 mL
0.0 mL
Initial NaOH
Buret Reading
Final NaOH
Buret Reading 12.2 mL
Trial 2
10.0 mL
12.2 mL
23.2 mL
Trial 3 Trial 4
10.0 mL 10.0 mL
23.2 mL 35.2 mL
35.2 mL 47.7 mL
79) Calculate the volume of NaOH solution used to neutralize 10.0 ml. of the standard HCl solution in trial 3 in the given diagram.
[Show your work.]

Answer 1

In trial 3, 10.0 mL of the 1.2 M HCl standard solution was used. We need to calculate the volume of NaOH solution used to neutralize this amount of HCl.

We can use the formula:

M1V1 = M2V2

where M1 is the molar concentration of the HCl solution, V1 is the volume of HCl solution used (10.0 mL), M2 is the molar concentration of the NaOH solution, and V2 is the volume of NaOH solution used to neutralize the HCl.

From the titration data, we can see that in trial 3, the initial buret reading of NaOH was 23.2 mL and the final buret reading was 35.2 mL. Therefore, the volume of NaOH solution used in trial 3 is:

35.2 mL - 23.2 mL = 12.0 mL

Now, we can plug in the values into the formula:

(1.2 M) x (10.0 mL) = (M2) x (12.0 mL)

Solving for M2, we get:

M2 = (1.2 M x 10.0 mL) / (12.0 mL) = 1.0 M

Therefore, the molar concentration of the NaOH solution is 1.0 M, and 12.0 mL of NaOH solution was used to neutralize 10.0 mL of the 1.2 M HCl solution in trial 3.