Question

10ml of a blood sample (containing calcium oxalate) is dissolved in acid. It required $20ml$ of $0.001M$ $KMn{O_4}$ (which oxidizes oxalate to carbon dioxide). Hence, $C{a^{2 + }}$ ion in $10ml$ blood is:
A. $0.200g$
B. $0.02g$
C. $2.00g$
D. $0.002g$

Answer 1

We must remember to determine the amounts or concentrations of substances or drugs present in the sample, chemists use a mix of chemical reaction and stoichiometric calculations in an exceedingly methodology known as quantitative analysis. Suppose, for instance, we all know the identity of a certain compound in a solution however not its concentration.

Complete step by step answer:
Given, amount of blood sample dissolved in acid = $10ml$
$0.001{\text{ }}M$ $KMn{O_4}$ = 20ml$$$M$ C{a^{2 + }}$ion in$10{\text{ }}ml$blood = ? The equation for the reaction is,$2KMn{O_4} + Ca{C_2}{O_4} + 4{H_2}S{O_4} \to 2MnS{O_4} + {K_2}S{O_4} + CaS{O_4} + 2C{O_2} + 4{H_2}O$$
The permanganate ion moles are equal to = $0.020l \times 0.001M = 2 \times {10^{ - 5}}M$
The number of moles of ${C_2}{O_4}^{2 - }$ ions are $\dfrac{5}{2} \times 2 \times {10^{ - 5}} = 5 \times {10^{ - 5}}$
Number of moles of calcium ions is also equal to this.
So, the mass of calcium ion = $= 40 \times 5 \times {10^{ - 5}} = 0.002g$

So, the correct answer is Option D.

Additional information:
We know that there are three main problems involved in the determination of calcium in small amounts of blood, low in this element. As we know when preparing the solution that contains all of the calcium but doesn't contain organic matter. The second is that the precipitation of the calcium within the presence of such other elements as commonly occurs, while the third consists within the accurate estimation of this precipitated calcium. The third is the estimation of the precipitated calcium by accurate. So when determination is progress, overcome these problems to get the correct amount.

Note: Now we can discuss another method of calculation:
$10{\text{ }}ml$ of $C{a^{2 + }}$ ion (as calcium oxalate $\left( {{C_2}{O_4}^{2 - }} \right) = 20ml$ of 0.001M$$$$KMn{O_4}
$KMn{O_4}$ oxidizes oxalate to carbon dioxide,
${C_2}{O_4} \to 2C{O_2}$
Therefore,
$M = \dfrac{{20 \times 0.005}}{{2 \times 10}}$
Hence, Molarity of $C{a^{2 + }} = 5 \times {10^{ - 3}}M$
Therefore $0.005 = \dfrac{x}{{40 \times 1l}}$
On simplification we get,
$x = 0.2g$
So, $1000ml$ contains = 0.2g$$$C{a^{2 + }}$$ Then 10ml$sample contains,$\dfrac{{0.2}}{{1000}} \times 10g = 0.002g$