To determine the concentration of sodium hydroxide (NaOH), we can use the concept of moles and stoichiometry from the balanced chemical equation:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
From the balanced equation, we see that 1 mole of HCl reacts with 1 mole of NaOH.
- Calculate the moles of HCl:
The concentration (\(C\)) and volume (\(V\)) of HCl are given as:
- Concentration of HCl: 0.10 M
- Volume of HCl: 40 mL = 0.040 L
Using the formula for moles (\(n = C \cdot V\)): \[ n_{\text{HCl}} = 0.10 , \text{M} \times 0.040 , \text{L} = 0.004 , \text{moles} \]
- Calculate the moles of NaOH:
Since the reaction shows a 1:1 ratio: \[ n_{\text{NaOH}} = n_{\text{HCl}} = 0.004 , \text{moles} \]
- Determine the concentration of NaOH:
We know the volume of NaOH used is 50 mL, which is equal to 0.050 L. We can now find the concentration (\(C_{\text{NaOH}}\)):
Using the moles we calculated: \[ C_{\text{NaOH}} = \frac{n_{\text{NaOH}}}{V_{\text{NaOH}}} \] \[ C_{\text{NaOH}} = \frac{0.004 , \text{moles}}{0.050 , \text{L}} = 0.08 , \text{M} \]
Thus, the concentration of sodium hydroxide is 0.08 M.
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