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telo118 [61]
3 years ago
12

A 20.0 milliliter sample of 0.200 molar K₂CO₃ solution is added to 30.0 milliliters of 0.400 molar Ba(NO₃)₂ solution. Barium car

bonate precipitates. The concentration of barium ion, Ba₂, in solution after the reaction is _________.
Chemistry
1 answer:
sveticcg [70]3 years ago
6 0

Answer:

0.32M

Explanation:

<u>Step 1:</u>  Balance the reaction

K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3

We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution

SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3

<u>Step 2: </u>Calculate concentration

To find the concentration of the barium cation we use the following equation:

Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>

<u />

<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3

[Ba2+] = 0.32 M

The concentration of Barium ion in solution is 0.32 M

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