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What are the concentration of the ions if 1.0 moles Al2(CO3)3 is dissolved in 0.50 L of solution? [Al3+] = 4.0 M &...

What are the concentration of the ions if 1.0 moles Al2(CO3)3 is dissolved in 0.50 L of solution?

[Al3+] = 4.0 M              [CO32-]= 6.0 M

[Al3+] = 2.0 M              [CO32-] = 3.0 M

[Al3+]= 1.5 M             [CO32-]= 1.5 M

[Al3+] = 12 M            [CO32-] = 18 M
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Answer #1

Answer:

Step 1: Calculate the Molarity (concentration) of a Al2(CO3)3

we know, Molarity = moles / volume (in L ) = 1 moles / 0.5 L = 2 M

Step 2: Calculation

The dissolution equation for Al2(CO3)3 in aqueous solution is:

Al2(CO3)3--------------> 2 Al+3 + 3 CO-23

Thus, every mole of Al2(CO3)3 dissolves into 2 moles of Al+3   and 3 mole of CO-23

If the original concentration of   Al2(CO3)3 is 2 M, then

[Al+3] = 2×2 M = 4 M

[ CO-23] = 3×2 M = 6 M

Hence, option(b)

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