Explanation:
The given data is as follows.
\(M_{1}\) = 0.6 M, \(V_{1}\) = ?
\(M_{2}\) = 0.5 M, \(V_{2}\) = 100.0 ml
Therefore, we calculate the volume of NaOH as follows.
\((Molarity \times Volume)_{HCl}\) = \((Molarity \times Volume)_{NaOH}\)
\(M_{1} \times V_{1} = M_{2} \times V_{2}\)
\(V_{1} = \frac{M_{2} \times V_{2}}{M_{1}}\)
= \(\frac{0.5 M \times 100.0 ml}{0.6 M}\)
= 83.33 ml
Thus, we can conclude that volume of HCl is 83.33 ml.
