Respuesta :
Answer
Explanation
Given that:
Mass of nickel (II) bromide (NiBr₂) that dissolved = 24.9 g
The volume of the aqueous solution of potassium carbonate (K₂CO₃) = 300 mL = 0.300 L
The molarity of K₂CO₃ = 0.60 M
What to find:
The final molarity of nickel(II) cation in the solution.
Step-by-step solution:
The first step is to write the balanced chemical equation for the reaction.
NiBr₂ + K₂CO₃ → NiCO₃ + 2KBr
The next step is to calculate the moles of NiBr₂ using:
[tex]\begin{gathered} Moles\text{ }(n)=\frac{mass\text{ }(m)}{M\text{ }(Molar\text{ }mass)} \\ \\ Molar\text{ }mass\text{ }of\text{ }NiBr₂=218.53\text{ }g\text{ /}mol \\ \\ Moles\text{ }of\text{ }NiBr₂=\frac{24.9g}{218.53g\text{/}mol}=0.1139\text{ }mol \end{gathered}[/tex]Also, the moles of K₂CO₃ can be calculated using:
[tex]\begin{gathered} Moles=Molarity\times Volume\text{ }in\text{ }L \\ \\ Moles\text{ }of\text{ }K₂CO₃=0.60M\times0.300L=0.1800\text{ }mol \end{gathered}[/tex]Using the mole ratio of NiBr₂ to K₂CO₃ from the equation above, that is (1:1)
we can say that K₂CO₃ is in excess.
So 0.1139 mol NiBr₂ gets converted into NiCO₃.
Hence, the final molarity of nickel(II) cation in the solution can be calculated using:
[tex]Molarity=\frac{Moles\text{ }of\text{ }Ni\text{ }in\text{ }solution}{Volume\text{ }of\text{ }solution}=\frac{0.1139mol}{0.300L}=0.380\text{ }M[/tex]Hence, the of nickel(II) cation in the solution is 0.380
