Respuesta :
Answer: The concentration of barium ions is 0.24 M
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .....(1)
- For barium nitrate:
Molarity of barium nitrate solution = 0.200 M
Volume of solution = 20 mL
Putting values in equation 1, we get:
[tex]0.200M=\frac{\text{Moles of barium nitrate}\times 1000}{20}\\\\\text{Moles of barium nitrate}=\frac{0.200\times 20}{1000}=0.04mol[/tex]
- For potassium carbonate:
Molarity of potassium carbonate solution = 0.400 M
Volume of solution = 30.0 mL
Putting values in equation 1, we get:
[tex]0.400M=\frac{\text{Moles of potassium carbonate}\times 1000}{30}\\\\\text{Moles of potassium carbonate}=\frac{0.400\times 30}{1000}=0.012mol[/tex]
The chemical equation for the reaction of potassium carbonate and barium nitrate follows:
[tex]K_2CO_3(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaCO_3(s)+2KNO_3(aq.)[/tex]
By Stoichiometry of the reaction:
1 mole of potassium carbonate reacts with 1 mole of barium nitrate
So, 0.012 moles of potassium carbonate will react with = [tex]\frac{1}{1}\times 0.012=0.012mol[/tex] of barium nitrate
As, given amount of barium nitrate is more than the required amount. So, it is considered as an excess reagent.
Thus, potassium carbonate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of potassium carbonate produces 1 mole of barium carbonate
So, 0.012 moles of potassium carbonate will produce = [tex]\frac{1}{1}\times 0.012=0.012mol[/tex] of barium carbonate
The chemical equation for the ionization of barium carbonate follows:
[tex]BaCO_3\rightarrow Ba^{2+}+CO_3^{2-}[/tex]
1 mole of barium carbonate produces 1 mole of barium ions and 1 mole of carbonate ions
Moles of barium ions = 0.012 moles
Volume of solution = (20 + 30) mL = 50 mL
Putting values in equation 1, we get:
[tex]\text{Concentration of }Ba^{2+}\text{ ions}=\frac{0.012\times 1000}{50}\\\\\text{Concentration of }Ba^{2+}\text{ ions}=0.24M[/tex]
Hence, the concentration of barium ions is 0.24 M
