Question

10. Suppose that 20.0 mL of 5.00 x 10-3 M aqueous sodium hydroxide (NaOH) is required to neutralize 10.0

Answer 1

Explanation:

B. 1.00 x 10^-2 M HCI

The question is asking us to find the molarity/concentration of the acid (HCl).

We have been provided with;

• 20.0 mL of 5.0 x 10^-³ M NaOH
• 10.0 mL of HCl

We know that molarity (M) of a solution is contained in 1 L or 1000 mL or 1000 cm³.

This means that;

5.0 x 10^-³ moles is contained in 1000 mL.

is contained in 1000 mL.X mol is contained in 20.0 mL

$x \: mol \: = \frac{20 \times 5.0 \times {10}^{ - 3} }{1000} \\ = 0.0001 \: mol$

= 0.0001 moles

These 0.0001 moles is contained in 10 mL of HCl.

To find the molarity of the acid;

0.0001 moles is contained in 10 mL

x mol is contained in 1000 mL

$x \: mol \: = \frac{0.0001 \times 1000}{10} \\ = 0.01 \: mol$

= 0.01 M

Therefore the Concentration of the acid (HCl) is 0.01 M or 1.0 x 10^-2 M