MCQ
A solution of potassium ferrocyanide would contains ...... ions
- A$2$
- B$3$
- C$4$
- ✓$ 5$
✓
Answer
Correct option: D.
$ 5$
d
(d)Potassium ferrocynide ${K_4}[Fe\,{(CN)_6}]$ will ionize as ${K_4}[Fe\,{(CN)_6}]$
(d)Potassium ferrocynide ${K_4}[Fe\,{(CN)_6}]$ will ionize as ${K_4}[Fe\,{(CN)_6}]$
$\mathop {[Co\,{{(N{H_3})}_5}S{O_4}]\,Br}\limits_{0.02\,{\text{mole}}} + AgN{O_3} \to $$\mathop {[Co\,{{(N{H_3})}_5}.\,S{O_4}]\,N{O_3}}\limits_{0.02\,{\text{mole }}(y)}$
$+ AgBr$
$\mathop {[Co\,{{(N{H_3})}_5}B{r_2}]S{O_4}}\limits_{0.02\,{\text{mole}}} + BaC{l_2} \to $$\mathop {[Co\,{{(N{H_3})}_5}Br]\,C{l_2}}\limits_{0.02\,{\text{mole}}\,(z)} $
$+ BaS{O_4}$
On using one lit. solution, we will get $ 0.01$ $mole$ $y $ and $0.01$ $mole$ $z.$
$4{K^ + } + {[Fe\,\,{(CN)_6}]^{4 - }}$
So, it will give five ions in solution.
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