A20.0–milliliter sample of 0.200–molar k2co3 so­lution is added to 30.0 milliliters of 0.400–mo­lar ba(no3)2 solution. barium carbonate precipi­tates. the concentration of barium ion, ba2+, in solution after reaction is:

0.16 M

Explanation:

Considering:

Or,

Given :

For :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of :

Moles of  = 0.004 moles

For :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 30.0×10⁻³ L

Thus, moles of :

Moles of  = 0.012 moles

According to the given reaction:

1 mole of reacts with 1 mole of

So,

0.012 mole of reacts with 0.012 mole of

Available mole of = 0.004 mole

Limiting reagent is the one which is present in small amount. Thus, is limiting reagent. (0.004 < 0.012)

The formation of the product is governed by the limiting reagent. So,

1 mole of reacts with 1 mole of and gives 1 mole of

0.004 mole of reacts with 0.004 mole of and gives 0.004 mole of

Left moles of = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20 + 30 mL = 50 mL = 0.050 L

So,

Concentration of barium ion, , in solution after reaction is:-

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