3 years ago

Please help me with this chem question!!

1 answer:

6 0

Answer:

caca123

Explanation:

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sweet [91]

Answer : The maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

Explanation :

The solubility equilibrium reaction will be:

$Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-$

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ni^{2+}][CN^-]^2$

Now put all the given values in this expression, we get:

$3.0\times 10^{-23}=(0.220+s)\times (2s)^2$

$s=5.84\times 10^{-12}M$

Therefore, the maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

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