Respuesta :
Answer:
0.16 M
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Or,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
Given :
For [tex]K_2CO_3[/tex] :
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 [tex]K_2CO_3[/tex] :
[tex]Moles=0.200 \times {20.0\times 10^{-3}}\ moles[/tex]
Moles of [tex]K_2CO_3[/tex] = 0.004 moles
For [tex]Ba(NO_3)_2[/tex] :
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 [tex]Ba(NO_3)_2[/tex] :
[tex]Moles=0.400 \times {30.0\times 10^{-3}}\ moles[/tex]
Moles of [tex]Ba(NO_3)_2[/tex] = 0.012 moles
According to the given reaction:
[tex]Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3[/tex]
1 mole of [tex]Ba(NO_3)_2[/tex] reacts with 1 mole of [tex]K_2CO_3[/tex]
So,
0.012 mole of [tex]Ba(NO_3)_2[/tex] reacts with 0.012 mole of [tex]K_2CO_3[/tex]
Available mole of [tex]K_2CO_3[/tex] = 0.004 mole
Limiting reagent is the one which is present in small amount. Thus, [tex]K_2CO_3[/tex] is limiting reagent. (0.004 < 0.012)
The formation of the product is governed by the limiting reagent. So,
1 mole of [tex]K_2CO_3[/tex] reacts with 1 mole of [tex]Ba(NO_3)_2[/tex] and gives 1 mole of [tex]BaCO_3[/tex]
0.004 mole of [tex]K_2CO_3[/tex] reacts with 0.004 mole of [tex]Ba(NO_3)_2[/tex] and gives 0.004 mole of [tex]BaCO_3[/tex]
Left moles of [tex]Ba(NO_3)_2[/tex] = 0.012 - 0.004 moles = 0.008 moles
Total volume = 20 + 30 mL = 50 mL = 0.050 L
So,
Concentration of barium ion, [tex]Ba^{2+}[/tex], in solution after reaction is:-
[tex]Molarity=\frac{0.008}{0.050}\ M = 0.16\ M[/tex]
