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 a moles each of `Na_(2)CO_(3)` and `NaHCO_(3)` are present in 1 g of equimolar mixture of two compounds then,
`(a xx 106) +(a xx 84)=1" "…(1)`
where 106 is the molar mass of `Na_(2)CO_(3)` and 84 is the molar mass of `NaHCO_(3)`.
Solving eq. (1) `190a=1 or a=(1)/(190) = 5.26 xx 10^(-3)`
Reactions involved during neutralization are:
`underset("a mol")underset("1 mol")(Na_(2)CO_(3)) + underset("2a mol")underset("2 mol")(2HCl) to 2NaCl +H_(2)O +CO_(2)`
`underset("a mol")underset("1 mol")(NaHCO_(3)) +underset("a mol")underset("1 mol")(HCl) to NaCl+H_(2)O +CO_(2)`
`:.` Total number of moles of HCl used to neutralize a moles each of `Na_(2)CO_(3)` and `NaHCO_(3)`
`=2a+a=3a = 3xx 5.26 xx 10^(-3)"mol"=1.578xx10^(-2)"mol" `
`:. 1.578 xx 10^(-2)"mol" = 0.1"mol L"^(-1) xx V`
`V=(1.578xx10^(-2)"mol")/(0.1" mol L"^(-1))=1.578xx10^(-1)L=157.8mL`