Question

calculate the concentration (in molarity) of an naoh solution if 25.0 ml of the solution is needed to neutralize 16.3 ml of a 0.465 m hcl solution.
round your answer to 3 significant digits.
m

Explanation:

Step1: Write the neutralization reaction

The reaction between \(NaOH\) and \(HCl\) is \(NaOH + HCl=NaCl + H_{2}O\). The mole ratio of \(NaOH\) to \(HCl\) is \(1:1\), so \(n(NaOH)=n(HCl)\).

Step2: Calculate the moles of \(HCl\)

The formula for moles \(n = C\times V\) (where \(C\) is concentration and \(V\) is volume in liters). For \(HCl\), \(C_{HCl}=0.465\space M\) and \(V_{HCl}=16.3\space mL=16.3\times10^{- 3}\space L\). Then \(n(HCl)=C_{HCl}\times V_{HCl}=0.465\times16.3\times10^{-3}\space mol\).

$$n(HCl)=0.465\times16.3\times10^{-3}=7.5795\times10^{-3}\space mol$$

Step3: Calculate the concentration of \(NaOH\)

Since \(n(NaOH) = n(HCl)\) and \(V_{NaOH}=25.0\space mL = 25.0\times10^{-3}\space L\), and \(C=\frac{n}{V}\). Then \(C_{NaOH}=\frac{n(NaOH)}{V_{NaOH}}=\frac{n(HCl)}{V_{NaOH}}\).

$$C_{NaOH}=\frac{7.5795\times 10^{-3}}{25.0\times10^{-3}}=\frac{7.5795}{25.0}=0.30318\space M$$

Answer:

\(0.303\space M\)