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Lady_Fox [76]
2 years ago
6

The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

Chemistry
1 answer:
sweet [91]2 years ago
7 0

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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Explanation:

The given data is as follows.

       Mass of mixture = 0.3471 g

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This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence,   moles of NaOH in 97 ml = \frac{0.1090 \times 97}{1000}

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           10.573 \times 10^{-3} mol - 4.326 \times 10^{-3} mol      

                = 6.247 \times 10^{-3} mol

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                  \frac{6.247 \times 10^{-3}}{3}

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                    2.082 \times 10^{-3} mol \times 40 g/mol

                         = 83.293 \times 10^{-3} g

                         = 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore,       40 g NaOH = 122 g benzoic acid

So,            0.0832 g NaOH = \frac{122 g}{40 g} \times 0.0832 g

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Thus, we can conclude that mass % of benzoic acid is 73.10%.

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3 years ago
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