# Question #b226c

Apr 22, 2017

${K}_{\text{sp}} = 1.6 \times {10}^{-} 9$.............

#### Explanation:

$5 {C}_{2} {O}_{4}^{2 -} + 2 M n {O}_{4}^{-} + 16 {H}^{+} \rightarrow 2 M {n}^{2 +} + 10 C {O}_{2} \left(g\right) \uparrow + 8 {H}_{2} O$.

So, we works out moles of $\text{permanganate}$:

$= 0.001 \cdot m o l \cdot {L}^{-} 1 \times 4 \times {10}^{-} 3 L = 4 \times {10}^{-} 6 \cdot m o l$

And thus, by the stoichiometry, there were $\frac{5}{2} \times 4 \times {10}^{-} 6 \cdot m o l$ $\text{calcium oxalate}$, which were dissolved in $250 \cdot m L$ water under conditions of saturation, i.e. a concentration of.......

$\frac{\frac{5}{2} \times 4 \times {10}^{-} 6 \cdot m o l}{0.250 \cdot L} = 4.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$.

But, by specification, this represented a saturated solution.

And so ${K}_{\text{sp}} = \left[C {a}^{2 +}\right] \left[{C}_{2} {O}_{4}^{2 -}\right]$

$= {\left(4.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1\right)}^{2} = 1.6 \times {10}^{-} 9$