Question

A 0.7860-g sample of impure solid Na2CO3 is neutralized by 23.48 mL of 0.1082 M HCl. What percentage purity of Na2CO3 was contained in the sample? There are no acidic or basic impurities present in the sample

Answer 1

Percentage purity of
`Na_(2)CO_(3)` in sample is 17.13%

Step-by-step explanation:

Net Ionic equation:
`2H^(+)+CO_(3)^(2-)\rightarrow H_(2)O+CO_(2)`

So, two moles of
`H^(+)` neutralize 1 mol of
`CO_(3)^(2-)`

Alternatively, two moles of HCl neutralize 1 mol of
`Na_(2)CO_(3)`

Number of moles of HCl added =
`(0.1082)/(1000)* 23.48mol=0.002541mol`

So, 0.002541 moles of HCl neutralizes 0.0012705 moles of
`Na_(2)CO_(3)`

Molar mass of
`Na_(2)CO_(3)` = 105.99 g/mol

So, mass of
`Na_(2)CO_(3)` present in impure sample =
`(0.0012705* 105.99)g=0.13466g`

So percentage purity of
`Na_(2)CO_(3)` in sample =
`(0.13466)/(0.7860)* 100` % = 17.13%