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

Question

HCI + NaOH ----> NaCl + H2O
The concentration of sodium hydroxide is _____.
A. 0.04 M
B. 0.125 M
C. 0.10 M
D. 0.08 M

Answer 1

To determine the concentration of sodium hydroxide (NaOH), we can start with the stoichiometry of the neutralization reaction given:

\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the balanced equation, we see that hydrochloric acid (HCl) and sodium hydroxide (NaOH) react in a 1:1 molar ratio.

1. Calculate moles of HCl used: \[ \text{Moles of HCl} = \text{Concentration (M)} \times \text{Volume (L)} \] The concentration of HCl is 0.10 M and the volume used is 40 mL, which is 0.040 L: \[ \text{Moles of HCl} = 0.10 , \text{M} \times 0.040 , \text{L} = 0.004 , \text{moles} \]

2. Using the 1:1 ratio: Since the reaction ratio is 1:1, the moles of NaOH will also be: \[ \text{Moles of NaOH} = 0.004 , \text{moles} \]

3. Calculate the concentration of NaOH: Now we know the moles of NaOH and the volume of NaOH solution is 50 mL, which is 0.050 L: \[ \text{Concentration of NaOH (M)} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH (L)}} \] \[ \text{Concentration of NaOH} = \frac{0.004 , \text{moles}}{0.050 , \text{L}} = 0.08 , \text{M} \]

Thus, the concentration of sodium hydroxide is 0.08 M. Therefore, the answer is:

D. 0.08 M