To find the concentration of sodium hydroxide (NaOH), we start by using the stoichiometry of the neutralization reaction provided:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
This reaction shows a 1:1 molar ratio between hydrochloric acid (HCl) and sodium hydroxide (NaOH).
First, we can calculate the number of moles of hydrochloric acid used:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \] \[ \text{Moles of HCl} = 0.10 , \text{M} \times 0.040 , \text{L} = 0.004 , \text{moles} \]
Since the reaction has a 1:1 ratio, the moles of NaOH will also be 0.004 moles.
Next, we can find the concentration of NaOH. The volume of NaOH used is 50 mL, which is equal to 0.050 L.
Now we can calculate the concentration of NaOH:
\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH}} \] \[ \text{Concentration of NaOH} = \frac{0.004 , \text{moles}}{0.050 , \text{L}} = 0.08 , \text{M} \]
Thus, the concentration of sodium hydroxide is 0.08 M.
The correct answer is 0.08 M.
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