To find the concentration of sodium hydroxide (NaOH), we can use the concept of mole ratios and the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), which is represented by the equation:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
From the balanced equation, we see that the mole ratio of HCl to NaOH is 1:1.
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Calculate moles of HCl used: \[ \text{Moles of HCl} = \text{Concentration (M)} \times \text{Volume (L)} \] \[ \text{Moles of HCl} = 0.10, \text{M} \times 0.040, \text{L} = 0.004, \text{moles} \]
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Since the mole ratio is 1:1, the moles of NaOH will also be: \[ \text{Moles of NaOH} = 0.004, \text{moles} \]
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Calculate the concentration of NaOH: \[ \text{Concentration (M)} = \frac{\text{Moles}}{\text{Volume (L)}} \] \[ \text{Volume of NaOH} = 50, \text{mL} = 0.050, \text{L} \] \[ \text{Concentration of NaOH} = \frac{0.004, \text{moles}}{0.050, \text{L}} = 0.08, \text{M} \]
Therefore, the concentration of sodium hydroxide is 0.08 M.
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