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?

The mass of HCl present in 500 mL of acid solution is 36.5 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is HCl

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

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