Question

In today's experiment, Solutions A and B are prepared as follows.

Solution A: Solution B:
2.0 mL of 3.00 x 10-4 M bromcresol green 2.0 mL of 3.00 x 10-4 M bromcresol green
5.0 mL of 1.60 M acetic acid (HAc) 2.0 mL of 0.160 M sodium acetate (NaAc)
2.0 mL of 0.200 M KCl diluted to a total volume of 50 mL
diluted to a total volume of 50 mL
What is the molarity of HAc in Solution A?
A. 6.4x 10-3
B. 3.20 x 10-4
C. 8.0 x 10-3
D. 0.160
E. 1.60

Answer 1

D. 0.160

Step-by-step explanation:

The solution A is obtained adding 2.0mL of a solution of bromocresol green, 5.0mL of 1.60M HAc and 2.0mL of a solution of KCl. The solution is diluted to 50mL

That means the HAc is diluted from 5.0mL to 50.0mL, that is:

50.0mL / 5.0mL = 10 times.

And the final concentration of HAc must be:

1.60M / 10 times =

0.160M

Right answer is:

D. 0.160