Question

Whenof unknown were passed through a Jones reductor, molybdate ion $\mathbf{\left(}{\mathbf{MoO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\right)}$was converted into ${\mathbf{Mo}}^{\mathbf{3}\mathbf{+}}$. The filtrate required 16.43mLof $\mathbf{0}\mathbf{.}\mathbf{01033}{\mathbf{MKMnO}}_{\mathbf{4}}$to reach the purple end point.

${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{+}{\mathbf{Mo}}^{\mathbf{3}\mathbf{+}}\mathbf{\to }{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{MoO}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}$

A blank required 0.04mL. Balance the reaction and find the molarity of Molybdate in the unknown.

Short answer

The molarity of the molybdate is$c\left({\mathrm{Mo}}^{3+}3\right)=0.0113M$

Step-by-step solution

Definition of redox titration

• Redox reactions are oxidation-reduction chemical reactions in which the oxidation states of the reactants change. The term redox refers to the reduction-oxidation process.
• All redox reactions can be divided down into two types of reactions: reduction and oxidation.
• In a redox reaction, or Oxidation-Reduction process, the oxidation and reduction reactions always happen at the same time.

Find the net reaction.

To balance the equation we have to write two half-reactions so we can determine the ratio of moles

$$\begin{gathered} \mathrm{Reduction}:{\mathrm{MnO}}_4^{-}+5e^{-}+8H^{+}\rightleftharpoons M{n}^{2+}+4H_{2}O/\cdot 3 \\ \mathrm{Oxidation}:{\mathrm{Mo}}^{3+}+2H_{2}O\rightleftharpoons {\mathrm{MoO}}_2^{2+}+3e^{-}+4H^{+}/\cdot 5 \end{gathered}$$

To balance the number of electrons on the left and right side, we multiply the first reaction with 3 and the second reaction with 5 and we get:

$$\begin{gathered} \mathrm{Reduction}:3{\mathrm{MnO}}_4^{-}+15e^{-}+24H^{+}\rightleftharpoons 3{\mathrm{Mn}}^{2+}12H_{2}O \\ \mathrm{Oxidation}:5{\mathrm{Mo}}^{3+}+10H_{2}O\rightleftharpoons 5{\mathrm{MoO}}_2^{2+}+15e^{-}+20H^{+} \end{gathered}$$

By adding up these two reactions we get:

$3{\mathrm{MnO}}_4^{-}+15e+24H^{+}+5M{o}^{3+}+10H_{2}O\rightleftharpoons 3{\mathrm{Mn}}^{2+}+12H_{2}O+5{{\mathrm{MoO}}_2}^{2+}+15e+20H^{-}$

Net reaction is:

$3{{\mathrm{MnO}}_4}^{-}+5M{o}^{3+}+4H^{+}\rightleftharpoons 3{\mathrm{Mn}}^{2+}+5{{\mathrm{MoO}}_2}^{2+}+2H_{2}O$

Step 3:Find the molarity of the Molybdate.

To find the molarity of molybdate, we have to calculate the molarity of and put it in ratio with ${\mathrm{Mo}}^{3+}$. The data that is given:

$$\begin{gathered} \mathrm{V}(\mathrm{filtrate})=16.43\hspace{0.33em}\mathrm{mL} \\ \mathrm{V}(\mathrm{blank})=0.04\hspace{0.33em}\mathrm{mL} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=0.01033\hspace{0.33em}\mathrm{M} \\ \mathrm{V}\left(\mathrm{analyte}\right)=\mathrm{V}\left({\mathrm{Mo}}^{3+}\right)=25.00\hspace{0.33em}\mathrm{mL} \end{gathered}$$

We can calculate the volume of ${\mathrm{KMnO}}_4$by subtracting the volume of filtrate with the volume of blank:

$$\begin{gathered} \mathrm{V}\left({\mathrm{KMnO}}_4\right)=\mathrm{V}\left(\mathrm{filtrate}\right)-\mathrm{V}\left(\mathrm{blank}\right) \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=(16.43-0.04)\hspace{0.33em}\mathrm{mL} \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=16.39\hspace{0.33em}\mathrm{mL} \end{gathered}$$

The ratio of moles of ${\mathrm{MO}}^{3+}$and ${\mathrm{KMnO}}_4$and is 5;3 so we can write:

$$\begin{gathered} n\left({\mathrm{Mo}}^{3+}\right)=\frac{5}{3}\mathrm{n}\left({\mathrm{KMnO}}_4\right) \\ c\left({\mathrm{Mo}}^{3+}\right)\cdot V\left({\mathrm{Mo}}^{3+}\right)=\frac{5}{3}\cdot \mathrm{c}\left({\mathrm{KMnO}}_4\right)\cdot V\left({\mathrm{KMnO}}_4\right)/:V\left({\mathrm{Mo}}^{3+}\right) \\ c\left({\mathrm{Mo}}^{3+}\right))=\frac{5\cdot c\left({\mathrm{KMnO}}_4\right)\cdot \mathrm{V}\left({\mathrm{KMnO}}_4\right)}{3\cdot \mathrm{V}\left({\mathrm{Mo}}^{3+}\right)} \\ \mathrm{c}({\mathrm{Mo}}^{3+})=0.0113\hspace{0.33em}\mathrm{M} \end{gathered}$$

Hence, the molarity of the molybdate is$c\left({\mathrm{Mo}}^{3+}\right)=0.0113\hspace{0.33em}M$