Question

18.35 ml of an hcn solution was titrated with 35.4 ml of a 0.268 m naoh solution to reach the equivalence point. what is the molarity of the hcn solution? hcn + naoh → nacn + h₂o

Explanation:

Step1: Use the formula for titration at equivalence point

At equivalence point, \(n_{\text{HCN}}=n_{\text{NaOH}}\) (moles of acid = moles of base). The formula for moles \(n = C\times V\) (where \(C\) is molarity and \(V\) is volume in liters). So \(C_{\text{HCN}}\times V_{\text{HCN}}=C_{\text{NaOH}}\times V_{\text{NaOH}}\).

Step2: Convert volumes to liters

\(V_{\text{HCN}}=18.35\space mL = 18.35\times10^{- 3}\space L\), \(V_{\text{NaOH}}=35.4\space mL=35.4\times 10^{-3}\space L\), \(C_{\text{NaOH}} = 0.268\space M\).

Step3: Solve for \(C_{\text{HCN}}\)

From \(C_{\text{HCN}}\times V_{\text{HCN}}=C_{\text{NaOH}}\times V_{\text{NaOH}}\), we can express \(C_{\text{HCN}}=\frac{C_{\text{NaOH}}\times V_{\text{NaOH}}}{V_{\text{HCN}}}\).
Substitute the values: \(C_{\text{HCN}}=\frac{0.268\space M\times35.4\times 10^{-3}\space L}{18.35\times10^{-3}\space L}\)

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Answer:

\(0.517\space M\)