40g NaOH, 106g `Na_(2)CO_(3)` and 84g `NaHCO_(3)` is dissolved in water and the solution is made 1 litre. 20 mL of this stock solution is titrated with 1N HCl, hence which of the following statements is/are correct?

Options

A. the titre reading of HCl will be 40 mL, if phenolphthalein is used indicator from the very beginning
B. the titre reading of HCl will be 60 mL is phenolphthalein is used indicator from the very beginning
C. the titre reading of HCl will be 40 mL if the methyl orange is used indicator after the 1st end point
D. the tire reading of HCl will be 80 mL, if methyl orange is used as indicator from the very beginning

Answer

A, C, D

Explanation

`{:("Meq NaOH "= (40)/(40//1) xx 1000 = 1000),("Meq "Na_(2)CO_(3) = (106)/(106//2) xx 1000 = 2000),("Meq "NaHCO_(3) = (84)/(84//1) xx 1000 = 1000):}} ("meq" = (g)/(E) xx 1000) " in " 1.0 L`
In 20 ml solution, we have :
Meq `NaOH = 1000 xx(20)/(1000) = 20` Meq `Na_(2)CO_(3) = 2000 xx (20)/(1000) = 40`, Meq `NaOH CO_(3) = 1000 xx (20)/(1000) = 20`
(A) Meq `NaOH + (1)/(2) "Meq " Na_(2)CO_(3) = "Meq HCl "rArr (40)/(2) + 20 = 1 xx V_(HCl) rArr V_(HCl) = 40 ml`
`"Meq " Na_(2)CO_(2) "Left "[-=(1)/(2)("Meq "Na_(2)CO_(3) " initial")]+"Meq "NaHCO_(3) = "Meq HCl"`
`rArr (40)/(2) + 20 = 1 x V_(HCl) rArr V_(HCl) = 40 ml`
(D) Meq `NaOH + "Meq " Na_(2)CO_(3) + "Meq " NaHCO_(3) = "Meq HCl"`
`rArr 20+40+20 = 1 xx V_(HCl) rArr V_(HCl) = 80 ml`