How many ml of 0.1 HCl is required to react completely with 1.0g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equi-molar amounts of both ?

Explanation

Given 1gm mixture of `Na_(2)CO_(3)` and `NaHCO_(3)`
Let the mass of `Na_(2)CO_(3)=" a gms"`
Mass of `"NaHCO"_(3)=(1-a)" gms"`
Number of moles of `"Na"_(2)"CO"_(3)=("wt")/("GMW")=(a)/(106)`
Number of moles of `"NaHCO"_(3)=("wt")/("GMW")=(1-a)/(84)`
Given that the mixture contains Equi molar amounts of `"Na"_(2)"CO"_(3)` and `"NaHCO"_(3)`
`:. (a)/(106)=(1-a)/(84)`
`84" a"=106-106" a"`
`190"a"=106`
a = 0.558 gms
`:." Weight of "Na_(2)CO_(3)=0.558" gms"`
Weight of `"NaHCO"_(3)=1-0.558=0.442" gms"`
`"Na"_(2)"CO"_(3)+2" HCl"to2" NaCl"+"H"_(2)"O"+"CO"_(2)`
106 gms - 73 gms
`0.558"g"-?=(73xx0.558)/(106)=0.384" gms"`
`"NaHCO"_(3)+"HCl" to" NaCl"+"H"_(2)"O"+"CO"_(2)`
84 gms - 36.5 gms
0.442gms - ?
`=(36.5xx0.442)/(84)=0.1928`
`:." The weight of HCl required "=0.384+0.192=0.576" gms"`
Molarity (M) `=("Wt")/("GMWt")xx(1000)/("V"(ml))`
`0.1=(0.576)/(36.5)xx(1000)/("V")`
`V=(0.576xx1000)/(36.5xx0.1)=(576)/(3.65)=157.80" ml"`