Question

consider the reaction of 75.0 ml of 0.350 m c₅h₅n (kb = 1.7 x 10⁻⁹) with 100.0 ml of 0.425 m hcl.
what quantity in moles of c₅h₅n would be present before the reaction takes place?

Explanation:

Step1: Convert volume to liters

Volume of \(C_5H_5N\) solution \(V = 75.0\space mL=75.0\times10^{- 3}\space L\)

Step2: Use the formula \(n = M\times V\)

Molarity of \(C_5H_5N\) solution \(M = 0.350\space M\)
Number of moles \(n(C_5H_5N)=M\times V\)
\(n(C_5H_5N)=0.350\space mol/L\times75.0\times 10^{-3}\space L\)
\(n(C_5H_5N)=0.02625\space mol\)

Answer:

\(0.02625\space mol\)