Respuesta :
Answer:
0.2042 M is the original concentration of [tex]Ce^{4+}[/tex] (aq) in the titrating solution.
Explanation:
Mass of [tex]As_2O_3[/tex] = 0.217 g
Moles of [tex]As_2O_3=\frac{0.217 g}{198 g/mol}=0.001096 mol[/tex]
1 mole of [tex]As_2O_3[/tex] have 2 mole of As and 1 mole of [tex]H_3AsO_3[/tex] have 1 mole of As.
So, from 1 mole of [tex]As_2O_3[/tex] we will have 2 moles of [tex]H_3AsO_3[/tex]
Then from 0.001096 mol of [tex]As_2O_3[/tex] :
[tex]2\times 0.001096 mol=0.002192 mol[/tex] of [tex]H_3AsO_3[/tex]
[tex]2Ce^{4+}(aq)+H_3AsO_3(aq)+3H_2O(l)\rightarrow 2Ce^{3+}(aq)+H_3AsO_4(aq)+2H^+(aq)[/tex]
According to reaction, 1 mole of [tex]H_3AsO_3[/tex] reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of
[tex]\frac{2}{1}\times 0.002192 mol=0.004384 mol[/tex] of cerium (IV) ions.
Volume of the acidic cerium{IV) sulfate = 21.47 ml =0.02147 L
1 mL = 0.001 L
[tex]concentration = \frac{Moles}{Volume(L)}[/tex]
[tex][Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M[/tex]
0.2042 M is the original concentration of [tex]Ce^{4+}[/tex] (aq) in the titrating solution.
