The number of moles of solute dissolved per liter of solution; also known as molar concentration.

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

The soil of an agricultural field is too basic what can we do to correct it

katen-ka-za [31]

Answer:

you can correct it by adding an acid to make it neutral or less basic

3 0

3 years ago

What information is needed to calculate the average atomic mass of an element?(1 point)

g100num [7]

Answer:

Taking into account the definition of average atomic mass and isotopes of an element, the information that you need is the masses of its isotopes and their percent abundances.

Each chemical element is characterized by the number of protons in its nucleus, which is called the atomic number Z.

But in the nucleus of each element it is also possible to find neutrons, whose number can vary. The atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus.

The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.

The atomic mass of an element is the weighted average mass of its natural isotopes. Therefore, the atomic mass of an element is not a whole number.

The weighted average means that not all isotopes have the same percentage.

In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.

Explanation:

7 0

3 years ago

A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?

Shalnov [3]

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

5 0

3 years ago

The amount of matter in an object is its

andrezito [222]

The amount of matter in an object is its Mass...

7 0

3 years ago