`40 mL 0.05 M` solution of sodium sesquicarbonate dehydrate `(Na_(2)CO_(3).NaHCO_(3).2H_(2)O)` is titrated against `0.05 M HCl` solution, `x mL` of acid is required to reach the phenolphthalein end point while mL of same acid were required when methyl organe indicator was used in a separate titration. Which of the following is (are) correct statements?
a.`y-x = 80 mL`
b. `y+x = 160 mL`
c. If the titration is started with phenolphthalein indicator and methyl orange is added at the end point, `2 x mL` of `HCl` would be required further to reach the end point
d. If the same volume of same solution is titrated against `0.10 M NaOH, x//2 mL` of base would be required

Options

A. `80 mL
B. 30 mL
C. 120 mL
D. none

Answer

A

Explanation

0.05 M `Na_(2)CO_(3)=0.1 N Na_(2)CO_(3)`
`0.05 M NaHCO_(3)=0.05 N NaHCO_(3)`
`40 L of 0.1 Na_(2)CO_(3)=40 " mL of " 0.1 N HCl`
(For complete reaction) `=80" mL of " 0.05 M HCl`
For `50%` reaction`=40 " mL of " 0.05 M HCl`
With phenolphthalein `=x=40 mL`
`40 " mL of " 0.05 NaHCO_(3)=40 " mL of " 0.05 M HCl`
With methyl orange, `y=80+40=120mL`
`therefore(y-x)=80mL`