Question

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 carbonate precipi­tates. The concentration of barium ion, Ba2+, in solution after reaction is:

Answer 1

Answer:

0.16 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

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 $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$  = 0.004 moles

For $Ba(NO_3)_2$ :

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 $Ba(NO_3)_2$ :

$Moles=0.400 \times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$  = 0.012 moles

According to the given reaction:

$Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3$

1 mole of $Ba(NO_3)_2$ reacts with 1 mole of $K_2CO_3$

So,

0.012 mole of $Ba(NO_3)_2$ reacts with 0.012 mole of $K_2CO_3$

Available mole of $K_2CO_3$ = 0.004 mole

Limiting reagent is the one which is present in small amount. Thus, $K_2CO_3$ is limiting reagent. (0.004 < 0.012)

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

1 mole of $K_2CO_3$ reacts with 1 mole of $Ba(NO_3)_2$ and gives 1 mole of $BaCO_3$

0.004 mole of $K_2CO_3$ reacts with 0.004 mole of $Ba(NO_3)_2$ and gives 0.004 mole of $BaCO_3$

Left moles of $Ba(NO_3)_2$ = 0.012 - 0.004 moles = 0.008 moles

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

So,

Concentration of barium ion, $Ba^{2+}$, in solution after reaction is:-

$Molarity=\frac{0.008}{0.050}\ M = 0.16\ M$