Question

titration example: what volume (in ml) of a 0.245 m h₂so₄ solution is required to titrate 22.50 g of mg(oh)₂?

Explanation:

Step1: Calculate moles of Mg(OH)₂

The molar mass of Mg(OH)₂ is \(M = 24 + 2\times(16 + 1)=58\space g/mol\).
The number of moles of Mg(OH)₂, \(n=\frac{m}{M}\), where \(m = 22.50\space g\). So \(n=\frac{22.50}{58}\space mol\approx0.388\space mol\).

Step2: Write the chemical reaction equation

The reaction between \(H_2SO_4\) and \(Mg(OH)_2\) is \(H_2SO_4+Mg(OH)_2 = MgSO_4 + 2H_2O\). The mole ratio of \(H_2SO_4\) to \(Mg(OH)_2\) is \(1:1\).

Step3: Calculate moles of \(H_2SO_4\)

Since the mole ratio is \(1:1\), the number of moles of \(H_2SO_4\), \(n_{H_2SO_4}=n_{Mg(OH)_2}\approx0.388\space mol\)

Step4: Use the formula \(n = C\times V\) to find volume \(V\)

Given \(C = 0.245\space M\) (molarity of \(H_2SO_4\)), and \(n = 0.388\space mol\). From \(V=\frac{n}{C}\), we have \(V=\frac{0.388}{0.245}\space L\)
\(V=\frac{0.388}{0.245}\times1000\space mL\approx1580\space mL\)

Answer:

\(1580\space mL\)