It takes 40 mL of 0.10 M of hydrochloric acid (HCl) to neutralize 50 mL of sodium hydroxide (NaOH) for the following chemical reaction:

Question

HCl + NaOH → NaCl + H2O

The concentration of sodium hydroxide is _____.

(1 point)
Responses

0.08 M
0.08 M

0.04 M
0.04 M

0.125 M
0.125 M

0.10 M
0.10 M

Answer 1

To determine the concentration of sodium hydroxide (NaOH), we can start by using the information given about the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH).

The balanced chemical reaction is: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the balanced equation, we can see that the stoichiometry is 1:1, meaning 1 mole of HCl reacts with 1 mole of NaOH.

1. Calculate moles of HCl:

   • Volume of HCl = 40 mL = 0.040 L
   • Concentration of HCl = 0.10 M
   • Moles of HCl = Concentration × Volume = \( (0.10 , \text{mol/L}) \times (0.040 , \text{L}) = 0.0040 , \text{mol} \)

2. Because of the 1:1 stoichiometry, moles of NaOH = moles of HCl:

   • Moles of NaOH = 0.0040 mol

3. Calculate concentration of NaOH:

   • Volume of NaOH = 50 mL = 0.050 L
   • Concentration of NaOH = Moles/Volume = \( \frac{0.0040 , \text{mol}}{0.050 , \text{L}} = 0.080 , \text{M} \)

Thus, the concentration of sodium hydroxide (NaOH) is 0.08 M.