# Question 9cee6

Oct 10, 2016

I take it this is 2nd/3rd year university question?

#### Explanation:

Deep purple, potassium permanganate is reduced to colourless $M {n}^{2 +}$:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i\right)$

Charge and mass are balanced, so we know this is right.

And the $P {t}^{2 +}$ is oxidized to $P {t}^{4 +}$:

$P {t}^{2 +} \rightarrow P {t}^{4 +} + 2 {e}^{-}$ $\left(i i\right)$.

We wish to remove the electrons, so we take the cross product, $2 \times \left(i\right) + 5 \times \left(i i\right)$:

$2 M n {O}_{4}^{-} + 16 {H}^{+} + 5 P {t}^{2 +} \rightarrow 2 M {n}^{2 +} + 5 P {t}^{4 +} + 8 {H}_{2} O$

The which I tink is balanced with respect to mass and charge. So this equation tells us that 2 equiv permanganate would oxidize 5 equiv platinum ion. The endpoint of the titration would be signalled by the persistence of the striking purple colour of $M n {O}_{4}^{-}$ ion.

$26.87 \times {10}^{-} 3 \cdot L \times 0.0500 \cdot m o l \cdot {L}^{-} 1 = 1.344 \times {10}^{-} 3 \cdot m o l$ $K M n {O}_{4}$ were used. And thus $\left[P {t}^{2 +}\right]$ $=$ $\frac{\frac{5}{2} \times 1.344 \times \cancel{{10}^{-} 3} \cdot m o l}{250.0 \times \cancel{{10}^{-} 3} \cdot L}$ $=$ $1.344 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

If there were $250.0 \cdot m L$ of the starting platinum solution, there were $250.0 \times {10}^{3} L \times 1.344 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$ $=$ ??mol, ??g# $P {t}^{2 +}$?

I do not particularly like this question, because they should not be asking you questions about experiments that have not been performed in the practical. I have never oxidized $P {t}^{2 +}$ titrimetrically, and I doubt that the examiner has either.