Respuesta :
Answer:
The concentration of the potassium ions in the original potassium chromate solution is 4.2927 mol/L.
The concentration of the chromate ions in the final solution is 1.0731 mol/L.
Explanation:
[tex]K_2CrO_4+2AgNO_3\rightarrow Ag_2CrO_4+2KNO_3[/tex]
Volume of solution A i.e. solution of silver nitrate = 466.0 mL = 0.466 L
Volume of solution B i.e. solution of potassium chromate = 466.0 mL = 0.466 L
Moles of silver chromate =[tex]\frac{331.8}{331.73 g/mol}=1.0002 mol[/tex]
According to reaction , 1 mol of silver chromate is produce from 2 moles of silver nitrate.
Then, 1.0002 moles of silver chromate will be formed from:
[tex]\frac{1}{2}\times 1.0002 mol=0.5001 mol[/tex] of silver nitrate.
According to reaction , 1 mol of silver chromate is produce from 1 mole of potassium chromate.
Then, 1.0002 moles of silver chromate will be formed from:
[tex]\frac{1}{1}\times 1.0002 mol=1.0002 mol[/tex] of potassium chromate
[tex]Concentration =\frac{Moles}{Volume (L)}[/tex]
a) The concentration of the potassium ions in the original potassium chromate solution.
Volume of the original solution = 0.466 L
1 mol of potassium chromate dissociates into 2mol of potassium ions and 1 mol of chromate ions:
Moles of potassium ions = 2 × 1.0002 mol = 2.0004 mol
[tex][K^+]=\frac{2.0004 mol}{0.466 L}=4.2927 mol/L[/tex]
b) The concentration of the chromate ions in the final solution
Volume of the final solution = 0.466 L + 0.466 L
Moles of chromate ions = 1 × 1.0002 mol = 1.0002 mol
[tex][CrO_4^{2+}]=\frac{1.0002 mol}{0.466 L+0.466L}=1.0731 mol/L[/tex]
