How many mL of 0.1 M HCl are required to react completely with 1 g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of both ?

Explanation

Let the amount of `Na_(2)CO_(3)` in the mixture be `x_(g)`, Then, the amount of `NaHCO_(3)` in the mixture is `(1-x)g`.
Molar mass of `Na_(2)CO_(3)=2xx23+1xx12+3xx16`
`= 106 g mol^(-1)`
`therefore` Number of moles `Na_(2)CO_(3)=(x)/(106)mol`
Molar mass of `NaHCO_(3)=1xx23+1xx12+3xx16`
`= 84 g mol^(-1)`
Number of moles `NaHCO_(3)=(1-x)/(84)` mol
According to the question,
`(x)/(106)=(1-x)/(84)`
`therefore 84x = 106-106 x`
`therefore 190x=106`
`therefore x=0.5579`
Therefore, number of moles of `Na_(2)CO_(3)`
`=(0.5579)/(106)mol`
= 0.053 mol
And, number of moles of `NaHCO_(3)`
`= (1-0.5579)/(84)`
= 0.0053 mol
HCl reacts with `Na_(2)CO_(3)` and `NaHCO_(3)` according to the following equation.
`2HCl+Na_(2)CO_(3)to 2NaCl+H_(2)O+CO_(2)`
`HCl+NaHCO_(3)to NaCl+H_(2)O+CO_(2)`
1 mol of `Na_(2)CO_(3)` reacts with 2 mol of HCl.
Therefore, 0.0053 mol of `Na_(2)CO_(3)` reacts with `2xx0.0053` mol = 0.0106 mol of HCl.
Similarly, 1 mol of `NaHCO_(3)` reacts with 1 mol of HCl. Therefore, 0.0053 mol of `NaHCO_(3)` reacts with 0.0053 mol of HCl.
Total moles of HCl required `= (0.0106+0.0053)` mol
= 0.0159 mol
In 0.1 M of HCl
0.1 mol of HCl is preset in 1000 mL of the solution
Therefore, 0.0159 mol of HCl is present in
`= (1000xx0.0159)/(0.1)` mol
= 159 mL of the solution
Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of both.