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What is the amount of copper ions present in a 1.0 mL sample of copper sulfate solution, when a current of 3.0 A for 10 seconds is used for coulometry analysis, assuming 100% efficiency in electrode reactions and a coulombic efficiency of 90%?

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To determine the amount of copper ions present in the 1.0 mL sample of copper sulfate solution, we will first need to calculate the number of moles of copper ions reduced during the coulometry analysis.First, let's find the total charge  in coulombs  passed through the solution:Total charge  Q  = Current  I   Time  t Q = 3.0 A  10 s = 30 CSince the coulombic efficiency is 90%, the effective charge used for the reduction of copper ions is:Effective charge  Q_eff  = Total charge  Q   Coulombic efficiencyQ_eff = 30 C  0.90 = 27 CNow, we can use Faraday's constant  F  to determine the number of moles of copper ions reduced. Faraday's constant is approximately 96,485 C/mol e-. The balanced equation for the reduction of copper ions is:Cu + 2e  CuFrom the balanced equation, we can see that 2 moles of electrons are required to reduce 1 mole of copper ions. Therefore, we can calculate the number of moles of copper ions reduced  n_Cu  as follows:n_Cu =  Q_eff  /  2  F n_Cu = 27 C /  2  96,485 C/mol e- n_Cu = 27 C / 192,970 C/moln_Cu  1.40  10 molNow that we have the number of moles of copper ions, we can calculate the concentration of copper ions in the 1.0 mL sample of copper sulfate solution:Concentration  C  = n_Cu / Volume  V C = 1.40  10 mol / 1.0  10 LC  0.14 MTherefore, the concentration of copper ions in the 1.0 mL sample of copper sulfate solution is approximately 0.14 M.

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