8. Consider the following equilibrium: A solution is made by mixing 11.0 mL of 1

Question

8. Consider the following equilibrium:

A solution is made by mixing 11.0 mL of 1.10 M CuSO4 and 1.20 L of 0.300 M NH3. Assume additive volumes to answer the following questions. Give all answers to three sig figs.

a) What is the concentration of NH3 in the resulting solution?

b) What is the concentration of Cu(NH3)42+ in the resulting solution?

c) What is the concentration of Cu2+ in the resulting solution?

Consider the following equilibrium: A solution is made by mixing 11.0 mL of 1.10 M CuSO4 and 1.20 L of 0.300 M NH3. Assume additive volumes to answer the following questions. Give all answers to three sig figs. What is the concentration of NH3 in the resulting solution? What is the concentration of Cu(NH3)42+ in the resulting solution?

Explanation / Answer

This problem is laid out a little different from other types. Other types are dissociations of acids, bases, and low-solubility salts, and K = [products]/[reactants]. In this one, you have Kf,which is:

Kf = [Cu(NH3)4++]/[Cu2+][NH3]^4 = 4.8 x10^12

Let the CuSO4 solution be called CS. Let the NH3 solution be called AS.

20.0mLCS x 1..molCuSO4/1000mLCS = 0.02 mole CuSO4

20.0mLCS + 1400mLAS = 1420 mL total solution

0.02molCuSO4/1420mL x 1000mL/1L= 1.41 x 10^-2 mole CuSO4/1L

The 1.40 L of AS is so large that adding 20.0 mL CS to it does not change the concentration of NH3 that much, so neglect it.

Let [Cu2+] = x. Then [NH3] = 4x, and [Cu(NH3)4++] = 1.41 x 10^-2 - x. But x will turn out to be so small relative to 1.41 x 10^-2, that we can neglect that too. So 1.41 x 10^-2 is answer b).

(1.41x10^-2)/(x)(x)^4 = 4.8 x 10^12

x^5 = 2.94 x 10^-15; x = 1.24 x10^-3 = [Cu2+] which is answer c)

[NH3] = 4x = 4.96 x 10^-3

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