Objective question (single correct answer).
i. `H_(3) PO_(4)` is a tribasic acid and one of its salt is `NaH_(2) PO_(4)`. What volume of `1M NaOH` solution should be added to `12 g` of `NaH_(2) PO_(4)` to convert in into `Na_(3) PO_(4)` ?
a. `100 mL` b. 2 mol of `Ca (OH)_(2)` c. Both d. None
iii. The normality of a mixture obtained mixing `100 mL` of `0.2 m H_(2) SO_(4)` with `100 mL` of `0.2 M NaOH` is:
a. `0.05 N` b. `0.1 N` c. `0.15 N` d. `0.2 N`
iv `100 mL` solution of `0.1 N HCl` was titrated with `0.2 N` `NaOH` solutions. The titration was discontinued after adding `30 mL` of `NaOH` solution. The reamining titration was completed by adding `0.25 N KOH` solution. The volume of `KOH` required from completing the titration is:
a. `70 mL` b. `35 mL` c. `32 mL` d. `16 mL`

Answer

B, D

Explanation

a. b. `Mw "of" NaH_(2) PO_(4) = 120 g` (it is dibasic) (valency factor = 2)
Eq of `NaOH = Eq "of" NaH_(2)PO_(4)`
`(n = 1)` `(n = 2)`
`1 M xx 1 xx V_(L) = (12)/(120) xx 2`.
`V_(L) = 0.26 L = 200 mL`
ii.b. `SO_(2) Cl_(2) + 2H_(2) O rarr H_(2) SO_(4) + 2HCl`
`{:(1"mol" H_(2)SO_(4) -= 2NaOH),(2"mol"HCl -= 2NaOH):}] = 4 "mol" NaOH`
`{:(1"mol" H_(2)SO_(4) -= 1"mol" Ba(OH)_(2)),(2"mol"HCl -= 1"mol"Ba(OH)_(2)):}] implies {:(2"mol"),(Ba(OH)_(2)):}`
iii. b. `mEq "of" H_(2)SO_(4) = 100 xx 0.2 xx 1 = 20`
`mEq NaOH = 100 xx 0.2 xx 1 = 20`
`mEq "of" H_(2)SO_(4) "left" = 40 - 20 = 20`
Total volume `= 100 + 100 = 20`
`N` solution `= (20 mEq)/(200 mL) = 0.1 N`
iv. d. Total mEq of `HCl = 100 xx 0.1 = 10`
Total mEq of `NaOH = 300 xx 0.2 = 6`
mEq of `HCl` left `= 10 - 6 = 4`
mEq of `HCl = mEq "of" KOH`
`4 = 0.25 xx V, V implies 16 mL`