Question

25.0 ml of an lioh solution was titrated with 29.15 ml of a 0.205 m h3po4 solution to reach the equivalence point. what is the molarity of the lioh solution? h3po4 + 3lioh → li3po4 + 3h2o

Explanation:

Step1: Calculate the moles of \(H_3PO_4\)

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters).
\(V_{H_3PO_4}=29.15\ mL = 29.15\times10^{- 3}\ L\), \(C_{H_3PO_4}=0.205\ M\)
\(n_{H_3PO_4}=0.205\ M\times29.15\times 10^{-3}\ L=\frac{0.205\times29.15}{1000}=0.00597575\ mol\)

Step2: Determine the moles of \(LiOH\) using the stoichiometry

From the balanced equation \(H_3PO_4 + 3LiOH
ightarrow Li_3PO_4+3H_2O\), the mole ratio of \(H_3PO_4:LiOH = 1:3\)
\(n_{LiOH}=3\times n_{H_3PO_4}\)
\(n_{LiOH}=3\times0.00597575\ mol = 0.01792725\ mol\)

Step3: Calculate the molarity of \(LiOH\)

\(V_{LiOH}=25.0\ mL=25.0\times10^{-3}\ L\)
Use the formula \(C=\frac{n}{V}\)
\(C_{LiOH}=\frac{0.01792725\ mol}{25.0\times 10^{-3}\ L}=\frac{0.01792725}{0.025}=0.71709\ M\approx0.717\ M\)

Answer:

\(0.717\ M\)