Question

29.4 ml of a ch3cooh solution was titrated with 18.5 ml of a 0.175 m lioh solution to reach the equivalence point. what is the molarity of the ch3cooh solution? ch3cooh + lioh → ch3cooli + h2o

Explanation:

Step1: Calculate the moles of LiOH

Use the formula \(n = M\times V\) (where \(n\) is moles, \(M\) is molarity, \(V\) is volume in liters).
\(V_{LiOH}=18.5\space mL = 18.5\times10^{- 3}\space L\), \(M_{LiOH}=0.175\space M\)
\(n_{LiOH}=M_{LiOH}\times V_{LiOH}=0.175\space M\times18.5\times 10^{-3}\space L = 0.0032375\space mol\)

Step2: Relate moles of LiOH to moles of \(CH_3COOH\)

From the balanced chemical equation \(CH_3COOH + LiOH
ightarrow CH_3COOLi + H_2O\), the mole - ratio of \(CH_3COOH\) to \(LiOH\) is \(1:1\). So \(n_{CH_3COOH}=n_{LiOH} = 0.0032375\space mol\)

Step3: Calculate the molarity of \(CH_3COOH\)

\(V_{CH_3COOH}=29.4\space mL=29.4\times10^{-3}\space L\)
Use the formula \(M=\frac{n}{V}\)
\(M_{CH_3COOH}=\frac{n_{CH_3COOH}}{V_{CH_3COOH}}=\frac{0.0032375\space mol}{29.4\times 10^{-3}\space L}\)
\(M_{CH_3COOH}=\frac{0.0032375}{0.0294}\space M\approx0.110\space M\)

Answer:

\(0.110\space M\)