During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with

Question

4 years ago

5

1.0 M NaOH; 40. mL of the base were required to neutralize the sample. How many moles of acid are present in 2.0 liters of this unknown solution?

1 answer:

Liula [17] · Answer 1 · 2022-01-09T12:52:58+03:00

Answer:

8.0 moles

Explanation:

Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.

Using the formula: $\frac{concentration of acid X volume of acid}{concentration of base X volume of base} = \frac{mole of acid}{mole of base}$

Concentration of acid = ?

Volume of acid = 10 mL

Concentration of base = 1.0 M

Volume of base = 40 mL

mole of acid = 1

mole of base = 1

Substitute into the equation:

$\frac{concentration of acid X 10}{1.0 X 40} = \frac{1}{1}$

Concentration of acid = 40/10 = 4.0 M

To determine the number of moles of acid present in 2.0 liters of the unknown solution:

Number of moles = Molarity x volume

molarity = 4.0 M

Volume = 2.0 Liters

Hence,

Number of moles = 4.0 x 2.0 = 8 moles