During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200. mL of the base was required to neutralize the sample. How many grams of HCl are present in 500. mL of this acid solution
A) 46.5 g
B) 36.5 g
C) 26.5 g
D) 16.5 g
E) None of the above

Question

Quizplus · Accepted Answer

First, calculate the number of moles of NaOH used in the titration:
0.500 mol/L x 0.200 L = 0.100 mol NaOH

Since the reaction between HCl and NaOH is a 1:1 ratio, there are also 0.100 mol of HCl in the 50.0 mL sample used in the titration.

To calculate the concentration of HCl in the original solution, use the formula:
moles of solute/volume of solution in liters

0.100 mol HCl/0.050 L = 2.00 M HCl

Finally, to calculate the grams of HCl in 500. mL of the solution:
2.00 mol/L x 0.500 L = 1.00 mol HCl
1.00 mol HCl x 36.5 g/mol = 36.5 g

Therefore, the best choice is B) 36.5 g.

During a Titration the Following Data Were Collected