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What is the molar solubility of nickel(II) sulfide in 0.091 M KCN? For NiS, Ksp = 3.0 × 10 –19; for Ni(CN) 4 2–, Kf = 1.0 × 10 3

Marta_Voda [28]

Answer:

The value is $x = 0.0227 \ M$

Explanation:

From the question we are told that

The concentration of $KCN \ \ i.e \ \ CN^{-}$ is $M_1 = 0.091 \ M$

The solubility product constant for $NiS$ is $K_{sp} = 3.0 *10^{-19}$

The stability constant for $Ni(CN)_4 ^{2-}$ is $K_f = 1.0 *10^{31}$

Generally the dissociation reaction for NiS is

$Ni S \underset{}{\stackrel{}{\rightleftharpoons}} Ni^{2+} + S^{2-}$

Generally the formation reaction for $Ni(CN)_4 ^{2-}$ is

$4CN^- + N_i ^{2+} \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_{4}$

Combining both reaction we have

$4CN^ - + NiS \ \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_4 + S^{2-}$

Gnerally the equilibrium constant for this reaction is

$K_c = K_{sp} * K_f$

=> $K_c = 3.0 *10^{-19 } * 1.0 *10^{31}$

=> $K_c = 3.0*10^{12}$

Generally the I C E table for the above reaction is

$4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}$

initial [ I] 0.091 0 0

Change [C] -4x +x + x

Equilibrium [E ] 0.091 - 4x x x

Here is x is the amount in term of concentration that is lost by $CN^-$ and gained by $Ni(CN)_4 ^{2-}$ and $S^{2-}$

Gnerally the equilibrium constant for this reaction is mathematically represented as

$K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}$

=> $3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}$

=> $3.0*10^{12}* [0.091 - 4x ]^4 = x^2$

=> $[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}$

=> $[0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}$

=> $[0.091 - 4x ] = \frac{\sqrt{x} }{1316}$

=> $119.8 - 5264x =\sqrt{x}$

Square both sides

$(119.8 - 5264x)^2 =x$

=> $14352.04 - 1261255 x + 27709696x^2 = 0$

=> $27709696x^2 - 1261255 x + 14352.04 = 0$

Solving using quadratic equation

The value of x is $x = 0.0227 \ M$

Hence the amount in terms of molarity (concentration) of $Ni(CN)_4 ^{2-}$ and $S^{2-}$ produced at equilibrium is $x = 0.0227 \ M$ it then means that the amount of NiS (nickel(II) sulfide) lost at equilibrium is $x = 0.0227 \ M$

So the molar solubility of nickel(II) sulfide at equilibrium is

$x = 0.0227 \ M$

3 0

3 years ago

Balance the redox reaction in acid solution: I-(aq)>I3-(aq)+CI-(aq)

Vikentia [17]

Answer:

s

Explanation:

s

6 0

2 years ago

What is the ph value of water and salt?

ser-zykov [4K]

7 - Neutral because water and salt are neutral substances

4 0

4 years ago

a cylinder container is filled with air. its radius is 6cm and its height is 3.5cm. what is the volume of the air inside?

hram777 [196]

Answer:

45

Explanation:

I am math teacher

4 0

3 years ago

The following table lists molecular weight data for a polypropylene material. Molecular Weight Range (g/mol) xi wi

nikklg [1K]

Answer:

a) the number-average molecular weight is 32,400 g/mol

b) the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization is 770

Explanation:

Given the data in the question;

Molecular Mean Number Weight Number Weight

weight average average

(g/mol) Mi(g/mol) xi wi xiMi wiMi

8,000-16,000 12,000 0.07 0.03 840 360

16,000-24,000 20,000 0.15 0.09 3000 1800

24,000-32,000 28,000 0.26 0.21 7280 5880

32,000-40,000 36,000 0.27 0.27 9720 9720

40,000-48,000 44,000 0.18 0.28 7920 12320

48,000-56,000 52,000 0.07 0.12 3640 6240

∑ Mn=32,400 Mw=36,320

so;

a) the number-average molecular weight

Mn = ∑Mixi

so from the table above; summation of Row Mixi

Mn = ∑Mixi = 32,400

Therefore, the number-average molecular weight is 32,400 g/mol

b) the weight-average molecular weight

Mw = ∑Miwi

so from the table above; summation of Row Miwi

Mw = ∑Miwi = 36,320

Therefore, the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization

the degree of polymerization of polypropylene can be determined using number-average molecular and repeat unit molecular weight.

now, for polypropylene { CH₂ = CH - CH₃ }

the repeat unit consist of 3 carbon atoms and 6 hydrogen atoms

given that;

Atomic weight of Carbon mC = 12.01 g/mol and

Atomic weight of Hydrogen mH = 1.008 g/mol

now we find the repeat unit molecular weight of polypropylene

m = nCmC + nHmH

where n is the number of repeat of atoms

so we substitute

m = ( 3 × 12.01) + ( 6 × 1.008)

m = 36.03 + 6.048

m = 42.078 g/mol

now we calculate the degree of polymerization;

DP = Mn / m

so we substitute

DP = 32,400 / 42.078

DP = 769.9985 ≈ 770

Therefore, the degree of polymerization is 770

3 0

3 years ago