Answer:
Percentage purity of [tex]Na_{2}CO_{3}[/tex] in sample is 17.13%
Explanation:
Net Ionic equation: [tex]2H^{+}+CO_{3}^{2-}\rightarrow H_{2}O+CO_{2}[/tex]
So, two moles of [tex]H^{+}[/tex] neutralize 1 mol of [tex]CO_{3}^{2-}[/tex]
Alternatively, two moles of HCl neutralize 1 mol of [tex]Na_{2}CO_{3}[/tex]
Number of moles of HCl added = [tex]\frac{0.1082}{1000}\times 23.48mol=0.002541mol[/tex]
So, 0.002541 moles of HCl neutralizes 0.0012705 moles of [tex]Na_{2}CO_{3}[/tex]
Molar mass of [tex]Na_{2}CO_{3}[/tex] = 105.99 g/mol
So, mass of [tex]Na_{2}CO_{3}[/tex] present in impure sample = [tex](0.0012705\times 105.99)g=0.13466g[/tex]
So percentage purity of [tex]Na_{2}CO_{3}[/tex] in sample = [tex]\frac{0.13466}{0.7860}\times 100[/tex] % = 17.13%
