Question

Calculate the concentration of nickel(II) ion in solution after the addition of 25.0 mL of 0.200 M NaCN to 60.0 mL of 0.0100 M Ni(NO3)2. (Kf is 2.0 × 1031 for [Ni(CN)4] 2─)

Answer 1

Solution :-

moles of NaCN = molarity * volume in liter   

                          = 0.200 mol per L * 0.025 L

                          = 0.005 mol

moles of Ni(NO3)2 = 0.0100 mol per L * 0.060 L = 0.0006 mol

now lets calculate the moles of the CN- needed to react with Ni2+

0.0006 mol Ni2+ * 4 mol CN- / 1 mol Ni^2+ = 0.0024 mol CN-

moles of CN- are more than needed for the reaction

so all of the Ni^2+ is converted to the Ni(CN)4^2-

therefore moles of the Ni(CN)4^2- that can be produced = 0.0006 mol

molarity of the Ni(CN)4^2- at total volume is

total volume = 60.0 ml + 25.0 ml = 85.0 ml = 0.085 L

[Ni(CN)4^2-] = 0.0006 mol / 0.085 L = 0.007059 M

now using the Kf we can calculate the concnetration of the Ni^2+

Ni(CN)4^2-   -------- > Ni^2+   +4CN^-

0.007059                      0            0

-x                                 +x           +4x

0.007059-x                  x               4x

1/kf = [Ni^2+][CN^-]^4/[Ni(CN)4^2-]

1/2.0*10^31 = [x][4x]^4/[0.007059-x]

(1/2.0*10^31)*(0.007059-x) = 256x^5

solving for the x we get

x= 6.73*10^-8 M

So the concnetration of the Ni^2+ after the reaction is 6.73*10^-8 M