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What is the concentration of copper ions in a 25 mL sample of a 0.05 M copper sulfate solution when the sample is analyzed using coulometry with a current of 2.5 A for 10 seconds?

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To solve this problem, we need to determine the amount of copper ions reduced during the coulometry process and then calculate the new concentration of copper ions in the solution.First, let's find the amount of charge passed through the solution during the coulometry process:q = I  twhere q is the charge, I is the current, and t is the time.q = 2.5 A  10 s = 25 CoulombsNow, we need to find the number of moles of electrons involved in the reduction of copper ions:Copper sulfate  CuSO  dissociates into copper ions  Cu  and sulfate ions  SO  in the solution. The reduction of copper ions involves the following reaction:Cu + 2e  CuFrom the reaction, we can see that 2 moles of electrons are required to reduce 1 mole of copper ions.Next, we need to find the number of moles of electrons  n  passed through the solution:n = q / Fwhere F is the Faraday constant  96,485 C/mol .n = 25 C / 96,485 C/mol  0.000259 moles of electronsNow, we can find the number of moles of copper ions reduced:0.000259 moles of electrons   1 mole of Cu / 2 moles of electrons   0.0001295 moles of CuInitially, we had 0.05 M copper sulfate in 25 mL of solution. To find the initial moles of copper ions:moles of Cu = M  Vwhere M is the molarity and V is the volume in liters.moles of Cu = 0.05 mol/L  0.025 L = 0.00125 molesNow, we can find the new moles of copper ions after the reduction:0.00125 moles - 0.0001295 moles  0.0011205 molesFinally, we can calculate the new concentration of copper ions in the solution:New concentration = moles of Cu / volume in litersNew concentration = 0.0011205 moles / 0.025 L  0.04482 MThe concentration of copper ions in the 25 mL sample of the 0.05 M copper sulfate solution after the coulometry process is approximately 0.04482 M.

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