Question

2. An industrial chemist converts Zn(H2O)42+ to the more stable Zn(NH3)42+ by mixing 50.0 L of 0.0020 M Zn(H2O)42+ and 25.0 L of 0.15 M NH3. What is the final [Zn(H2O)42+]? Kf of Zn(NH3)42+ is 7.8 × 108.

Answer 1

To find the final concentration of Zn(H2O)42+, we use the concept of equilibrium constant (Kf) and stoichiometry. The final concentration is determined by the moles of the reactants and the total volume. In this case, the final concentration of Zn(H2O)42+ is 0 M.

Step-by-step explanation:

To determine the final concentration of Zn(H2O)42+ after mixing, we need to use the concept of equilibrium constant (Kf). The equation for the reaction is:

Zn(H2O)42+(aq) + 4NH3(aq) → Zn(NH3)42+(aq) + 4H2O(l)

First, let's calculate the moles of Zn(H2O)42+ and NH3:

Moles of Zn(H2O)42+ = concentration (M) x volume (L)

= 0.0020 M x 50.0 L

= 0.10 moles

Moles of NH3 = concentration (M) x volume (L)

= 0.15 M x 25.0 L

= 3.75 moles

Since the stoichiometry of the reaction is 1:4 between Zn(H2O)42+ and NH3, the limiting reactant is Zn(H2O)42+ because there are fewer moles of it.

Therefore, all the Zn(H2O)42+ will be converted to Zn(NH3)42+.

Finally, we can calculate the final concentration of Zn(H2O)42+:

Final concentration = moles of Zn(H2O)42+ / total volume

= 0 moles / (50.0 L + 25.0 L)

= 0 M