Question

Cerium (IV) ions are strong oxidizing agents in acid ic solution, oxidizing arsenio us acid to arsen ic acid accor ding to the following equa tion:
2Ce4+(aq)+H3AsO3(aq)+3H2O(l)→2Ce3+(aq)+H3AsO4(aq)+2H+(aq)
A sample of As2O3 weighing 0.217 g is dissolved in basic solution and then acidified to make H3AsO3. Its titration with a solution of acidic cerium{IV) sulfate requires 21.47 ml. Determine the original concentration of Ce^4+(aq) in the titrating solution

Answer 1

Answer:

0.2042 M is the original concentration of $Ce^{4+}$ (aq) in the titrating solution.

Explanation:

Mass of $As_2O_3$ = 0.217 g

Moles of $As_2O_3=\frac{0.217 g}{198 g/mol}=0.001096 mol$

1 mole of $As_2O_3$ have 2 mole of As  and 1 mole of $H_3AsO_3$ have 1 mole of As.

So, from 1 mole of $As_2O_3$ we will have 2 moles of $H_3AsO_3$

Then from 0.001096 mol of $As_2O_3$ :

$2\times 0.001096 mol=0.002192 mol$ of $H_3AsO_3$

$2Ce^{4+}(aq)+H_3AsO_3(aq)+3H_2O(l)\rightarrow 2Ce^{3+}(aq)+H_3AsO_4(aq)+2H^+(aq)$

According to reaction, 1 mole of $H_3AsO_3$ reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of

$\frac{2}{1}\times 0.002192 mol=0.004384 mol$ of cerium (IV) ions.

Volume of the  acidic cerium{IV) sulfate = 21.47 ml =0.02147 L

1 mL = 0.001 L

$concentration = \frac{Moles}{Volume(L)}$

$[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M$

0.2042 M is the original concentration of $Ce^{4+}$ (aq) in the titrating solution.