Final answer:
Using the relationship between pOH and pH along with given equilibrium concentrations, the ratio of {CuOH+}/{Cu2+} in a pH 7 solution is approximately 2:1.
Step-by-step explanation:
To answer the question regarding the ratios of {CuOH+}/{Cu2+} and {NH4+}/{NH3}, we use pH and the equilibrium constants for the reactions involved. Assuming a pH of 7 for the solution, the concentration of hydroxide ions ([OH-]) can be calculated using the relationship pOH = 14 - pH, which gives us a pOH of 7, and thus an [OH-] of 1 × 10-7 M. Using the provided information that the concentration of Cu2+ and OH- at equilibrium is 1.765 × 10-7 M and 3.530 × 10-7 M respectively, we can determine that the ratio of {CuOH+}/{Cu2+} is equal to (3.530 / 1.765), or about 2:1.