Calculate the concentrations of the constituent ions of .0112M CaCl2.
The answers are supposed to be [Ca+2]=.0155M [Cl-]=.310M, but I keep getting [Ca+2]=.0112M and [Cl-]=.0224M.
How should I do this problem?
Thank you! ^-^
2007-03-28 15:49:54 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Stick to your guns. The books are sometimes wrong.
2007-03-28 15:56:08 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
MW of Na2SO4 (2*23 + 32 + 4*sixteen) = 142. to that end 142 g equals one mole One mole dissolved in a single litre answer is a million M Now the ionic equation is two Na^+ + SO4^ -2. to that end, 2 equivalents of sodium ions required to combine with one equivalent of sulfate ion. for this reason, the concentration of sodium ions would be two times in molar words while in comparison with sulfate ions. to that end, one mole of sulfate ions will combine with 2 moles of sodium ions to variety a independent salt. you additionally can get the respond in seconds in case you wreck down the equation as follows Na2SO4 = 2 Na+ + SO4^ -2 to that end 2 moles of sodium ions combine with one mole of sulfate ions to variety a million mole of sodium sulfate
2016-12-19 15:57:53 · answer #2 · answered by inkeles 3 · 0⤊ 0⤋
The way you state the problem, you are solving it correctly.
2007-03-28 15:57:55 · answer #3 · answered by steve_geo1 7 · 0⤊ 0⤋