Question

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

mL of an aqueous solution of hydrochloric acid (HCI). Using the balanced chemical equation below,
find the molarity of the HCI solution?
NaOH (aq) + HCI (aq) → H₂O (1) + NaCl (aq)
A. 1.00 x 103 M HCI
B. 1.00 x 102 M HCI
C. 5.00 x 103 M HCI
D. 5.00 x 10-2 MHCI

Answer 1

Step-by-step 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 * 5.0 * {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 \: = (0.0001 * 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