Question

A solution is made that is 1.1×10−3 M in Zn(NO3)2 and 0.140 M in NH3.

Part A
After the solution reaches equilibrium, what concentration of Zn2+(aq) remains?
Express your answer using two significant figures.

Answer 1

Data given : Molarity of Za+2 (Zn(NO3)2) = 1.1*10^-3 M

                 Molarity of ammonia = 0.140 M

Kf from standard reference = 4.1 *10⁸

The equation can be written as ,

Zn⁺²   + 4 NH₃ ------> Zn(NH₃)₄

The Kf equation for the above reaction will be,

Now let us recall the equation,

Zn⁺²         +    4 NH₃          ------> Zn(NH₃)4

1.1*10^-3        0.140                 -

1.1*10^-3-x          0.140-4x                       x

Substituting the concentration in the Kf equation ,

Substituting the value of Kf,

Since this will be become a 5 power equation (x⁵) which we cannot solve normally, so we

will make an assumption that 'x' is very very small compared to 0.140 M

We will re check this assumption once we obtain the value of x

Applying this approximation in the Kf equation,

Simplifying the above equation

x = 4.1 *10⁸ * (1.1*10^-3 -x) * 0.140⁴

x = ( 4.1 *10⁸ *1.1*10^-3 * 0.140⁴ ) - ( 4.1 *10⁸ * 0.140⁴ )x

(1+( 4.1 *10⁸ * 0.140⁴ ))x = ( 4.1 *10⁸ *1.1*10^-3 * 0.140⁴ )

x= 1.0999*10^-3

Hence amount of Zn²⁺ remaining = 1.1*10-3 -1.0999*10^-3 =6.9838 * 10^-9 M

Also we can see that 1.0999*10^-3 is very small compared to 0.150 hence the assumption is justifiable,

After equilibrium has been reached, Concentration of Zn2+ remaining is 6.9838 * 10^-9 M