Question

use the references to access important values if needed for this question.an aqueous solution of hydrochloric acid is standardized by titration with a 0.103 m solution of calcium hydroxide.if 11.6 ml of base are required to neutralize 18.9 ml of the acid, what is the molarity of the hydrochloric acid solution?m hydrochloric acid

Explanation:

Step1: Write the balanced chemical equation

$$2HCl + Ca(OH)_2=CaCl_2 + 2H_2O$$
From the equation, the mole ratio of \(HCl\) to \(Ca(OH)_2\) is \(n_{HCl}:n_{Ca(OH)_2}=2:1\)

Step2: Calculate the moles of \(Ca(OH)_2\)

Use the formula \(n = C\times V\) (where \(C\) is concentration and \(V\) is volume in liters).
\(V_{Ca(OH)_2}=11.6\space mL=11.6\times10^{- 3}\space L\), \(C_{Ca(OH)_2}=0.103\space M\)
\(n_{Ca(OH)_2}=C_{Ca(OH)_2}\times V_{Ca(OH)_2}=0.103\space M\times11.6\times 10^{-3}\space L = 1.1948\times10^{-3}\space mol\)

Step3: Calculate the moles of \(HCl\)

Since \(n_{HCl}=2n_{Ca(OH)_2}\)
\(n_{HCl}=2\times1.1948\times 10^{-3}\space mol = 2.3896\times10^{-3}\space mol\)

Step4: Calculate the molarity of \(HCl\)

\(V_{HCl}=18.9\space mL = 18.9\times10^{-3}\space L\)
Use the formula \(C=\frac{n}{V}\)
\(C_{HCl}=\frac{n_{HCl}}{V_{HCl}}=\frac{2.3896\times 10^{-3}\space mol}{18.9\times10^{-3}\space L}\approx0.126\space M\)

Answer:

\(0.126\space M\)