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#1
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The concentration of Cu^2+ ions in the water(which also contains sulfate ions)discharged from certain industrial plant is determined by adding excess sodium sulfide(Na2S) solution to 0.800 L of the water. The molecular equation is; Na2S(aq) + CuSO4 (aq) --> CuS(s) + Na2SO4 (aq)
A. How do I write the ionic equation? (I can do it fine but I'm confused with the state of each... )B. How do I calculate the molar concentration of Cu^2+ in the water sample if 0.0177 g of solid CuS is formed? Is it true that the answer is 2,31 x 10^-4? My teacher once explained it to me but I wasn't paying attention. Pity me and I must be really bad at chemistry :x Oh yeah and I just remember that I have this question which I found during my visit to other school, here it is: Physiological saline used in intravenous drips is approximately 0,9 % NaCl solution. 100g of physiological saline solution was treated with excess AgNO3 solution and formed 2,22 of AgCl precipitate. Can you calculate the % mass of NaCl in the physiological saline solution. As for me, I can do up until the half way only... |
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#2
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You add Na2S to water containing Cu(2+)-ions and SO4(2-) ions. The net equation will therefore be:
Na2S + Cu(2+) --> Na(+) + CuS Now you have the stoichiometry and you have to convert the mass of CuS to mol. Use the stoichiometry to find the amount of Cu(2+) that is needed to form 0,0177 g CuS. The calculate molarity with c = n/V. |
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