Calculate the equivalent weight of KHC 2 O 4 when it is neutralised by KOH.

Question

Calculate the equivalent weight of KHC 2 O 4 when it is neutralised by KOH. (A) 128 (B) 64 (C) 42 (D) None of these

Tardigrade Admin · Accepted Answer

KHC 2 O 4 leftharpoons K ++ HC 2 O 4- leftharpoons H ++ C 2 O 42- KOH leftharpoons K ++ OH - There is one replaceable H in KHC 2 O 4, so its n - factor is 1. Its eq. wt =(128/1)=128.

Q. Calculate the equivalent weight of $KH C_{2} O_{4}$ when it is neutralised by KOH.

128

64

42

None of these

$KH C_{2} O_{4} ⇌ K^{+} + H C_{2} O_{4}^{-} ⇌ H^{+} + C_{2} O_{4}^{2 -}$

$K O H ⇌ K^{+} + O H^{-}$

There is one replaceable $H$ in $KH C_{2} O_{4},$ so its $n$ - factor is

1. Its eq. wt $= \frac{128}{1} = 128$ .

Q. Calculate the equivalent weight of KHC2​O4​ when it is neutralised by KOH.

128

64

42

None of these

KHC2​O4​⇌K++HC2​O4−​⇌H++C2​O42−​ KOH⇌K++OH− There is one replaceable H in KHC2​O4​, so its n - factor is 1. Its eq. wt =1128​=128.

Q. Calculate the equivalent weight of $KH C_{2} O_{4}$ when it is neutralised by KOH.

$KH C_{2} O_{4} ⇌ K^{+} + H C_{2} O_{4}^{-} ⇌ H^{+} + C_{2} O_{4}^{2 -}$

$K O H ⇌ K^{+} + O H^{-}$

There is one replaceable $H$ in $KH C_{2} O_{4},$ so its $n$ - factor is

1. Its eq. wt $= \frac{128}{1} = 128$ .