56.0 g KOH, 138.0 g `K_(2)CO_(3)` and 100.0 g `KHCO_(3)` is dissolved in water and the solution is made 1 L. 10 " mL of " this stock solution is titrated with 2.0 M HCl. Which of the following statements is/are correct?

Options

A. When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 60 mL
B. When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 40 mL.
C. When methyl orange is used as an indicator from the very beginning, the titre value of HCl will be 80 mL.
D. When methyl orange is used as an indicator after the first end point the titre value of HCl will be 30 mL.

Answer

B, C, D

Explanation

`m" Eq of "`(KOH)/(L)=(56)/((56)/(1))xx10^(3)=100=(20)/(20)mL` solution m" Eq of "`(K_(2)CO_(3))/(L)=(138)/((138)/(2))xx10^(3)=200=(40)/(20)mL` solution
`m" Eq of "(KHCO_(3))/(L)=(100)/((100)/(1))xx10^(3)=1000=(20)/(20)mL`
(b). With phenolphthalein:
`m" Eq of "KOH+(1)/(2)m" Eq of "K_(2)CO_(3)=m" Eq of "HCl`
`20+(40)/(2)=vxx1N(n=1)`
`V_(HCl)=40mL`
(c). With methyl orange:
`m" Eq of "KOH+m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl` ltBrgt `20+40+20=Vxx1N`
`V_(HCl)=80mL`
(d). With methyl orange after the first end point:
`(1)/(2)m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl`
`20+20=Vxx1N`
`V_(HCl)=40mL`