Asked by Kim
I stuck on this prob. Can you help me to do this?
10.00 mL of a solution containing 0.20 M NaCl and 0.25 M AlCl3 is introduced into a cation exchange column. How many moles of hydronium ions will be released by the column?
10.00 mL of a solution containing 0.20 M NaCl and 0.25 M AlCl3 is introduced into a cation exchange column. How many moles of hydronium ions will be released by the column?
Answers
Answered by
DrBob222
10 mL of 0.2 M NaCl contains 0.01 x 0.2M = 0.002 moles NaCl.
It contains 0.01 x 0.25 = 0.0025 moles AlCl3.
The column will release 0.002 moles for the Na^+ and 3*0.0025 moles for the Al^+3.
It contains 0.01 x 0.25 = 0.0025 moles AlCl3.
The column will release 0.002 moles for the Na^+ and 3*0.0025 moles for the Al^+3.
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