To neutralise completely 20 mL of 0.1 M aqueous solution of phosphrous acid `(H_(3)PO_(3))`, the volume of 0.1 M aqueous KOH solution required is :
A. 40 mL
B. 20 mL
C. 10 mL
D. 60 mL
Correct Answer - a
Since `H_(3)PO_(3)` is dibasic, 0.1 M solution =- 0.2 N solution.
`underset(("Acid"))(N_(1)V_(1))=underset(("Base"))(N_(2)V_(2))`
`(0.2 N)xx20 mL)=(0.1N)xxV_(2)`
`V_(2)=((0.2N)xx(20mL))/((0.1N))=40 mL.`
Correct Answer - B
`("No. of eq")_(NaOH)=("No. of eq.")_(H_(2)SO_(4))`
`rArr(1xx1)xxy=(0.6xx2)xx10`
`rArr y=12 ml`
`Now, ("No. of eq.")_(acid)=("No. of eq.")_(NaOH)`
`rArr Nxx5=(1xx1)xx12`
`rArr N=(12)/(5)=2.4`
Correct Answer - A
`H_(2)SO_(4)+2OH^(-)toSO_(4)^(2-)+2H_(2)O`
`H_(3)PO_(4)+OH^(-)toH_(2)PO_(4)^(-)+H_(2)O`
For same amount of alkali, `H_(2)SO_(4)` and `H_(3)PO_(4)` required are in the ratio 1:2 (by number of moles or by weight)
Correct Answer - C
`3 BaCl_(3)+2Na_(3)PO_(4)to Ba_(3)(PO_(4))_(2)+6NaCl`
Here `BaCl_(2)` is the limiting reactant
No. of moles of `Ba_(3)(PO_(4))_(2)` formed
`=(1)/(3)xx"N0. of moles of" BaCl_(2) = (2)/(3)` mol
The chemical equations involved in the reactions are :
`P_(4)O_(6) +6H_(2)O to 4H_(3)PO_(3)`
`(H_(3)PO_(3)+2NaOH to Na_(2)HPO_(4)+2H_(2)Oxx4)/(P_(4)O_(6)+8NaOH to 4 Na_(2)HPO_(4)+2H_(2)O)`
No. of moles of `P_(4)O_(6)` in 1.1 g `=("Mass of" P_(4)O_(6))/("Molar...
Correct Answer - a
Equivalent mass of oxalic acid
`=126/2=63`
Normally of exalic acid
`=6.3/63xx1000/250=0.4N`
`underset(("oxalic acid"))(N_(1)V_(1))=underset((NaOH))(N_(2)V_(2))`
`0.4xx10=0.1xxV_(2)`
`V_(2)=(0.4xx10)/(0.1)=40mL`
`[H_(2)SO_(4)]=(19.6)/(98)=0.2 M` and `KOH=(11.2)/(56)=0.2M`
so for equivalent point, `(0.2xx50xx2)/(0.2)=100mL` of `KOH`solution should be added.
`[H^(+)]=(0.2xx2-0.2)/(2)=0.1M` so `pH=1`.