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A solution of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ is standardized iodometrically titrated against $0.167g$ of $KBr{{O}_{3}}$. This process requires $50mL$ of the $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ solution. What is the normality of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$?

Last updated date: 20th Jun 2024
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Hint: In the standardisation of sodium thiosulphate, a known concentration of potassium borate is taken. As the disappearance of colour occurs, the volume of unknown is calculated and the concentration of sodium thiosulphate can be determined. At this endpoint, the moles of potassium borate is the same as the sodium thiosulphate.

Complete step by step answer:
In the iodometric titration of sodium thiosulphate with potassium borate, involving a redox reaction where indirect titration of iodine liberated occurs. Where the standard potassium borate is reacted with excess potassium iodide and sulphuric acid which generates iodine.
$BrO_{3}^{-}+6{{I}^{-}}+6{{H}^{+}}\to B{{r}^{-}}+3{{I}_{2}}+3{{H}_{2}}O$ (1)
Then, the sodium thiosulphate solution titrates the iodine generated producing a pale-yellow solution to which starch indicator is added producing a blue colour. On further titration with sodium thiosulphate, disappearance of blue colour occurs and the concentration of sodium thiosulphate used is calculated.
$\operatorname{Re}duction:\,{{I}_{2}}+2{{e}^{-}}\to 2{{I}^{-}}$
$Oxidation:\,2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2N{{a}^{+}}+2{{e}^{-}}$
$Overall\,\,reaction:\,{{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI$ (2)
Adding equation (1) and 3$\times$(2) and further balancing it, we get,
\[6N{{a}_{2}}{{S}_{2}}{{O}_{3}}+BrO_{3}^{-}+6{{I}^{-}}\text{+}6{{H}^{+}}\to \text{ }3N{{a}_{2}}{{S}_{4}}{{O}_{6}}+B{{r}^{-}}\text{+}6NaI+3{{H}_{2}}O\]
From the above equation, it is seen that $BrO_{3}^{-}$ reduces to $B{{r}^{-}}$ with the gain of six electrons as oxidation state of bromines changes from $\text{(+5) to (-1)}$.
$Equivalent\,weight\,of\,KBr{{O}_{3}}=\dfrac{molar\,mass}{numbers\,of\,electrons\,gained}$ $=\dfrac{167}{6}$
where, Molar mass of $KBr{{O}_{3}}=39+79.9+16\times 3=166.9\approx 167g$.
$Normality\,of\,N{{a}_{2}}{{S}_{2}}{{O}_{3}}=\dfrac{gram\,equivalent\,of\,KBr{{O}_{3}}}{volume\,(in\,L)}=\dfrac{given\,mass}{eq.wt\,\,\times \,\,volume}$
$=\dfrac{0.167\times 6}{167\times 50\times {{10}^{-3}}}=0.12N$

Therefore, normality of 50 mL of sodium thiosulphate used to titrate against 0.167 g of potassium borate is option (B)- 0.12 N.

Note: During this redox reaction, the number of electrons produced is equal to the n-factor, using which the normality or molarity of unknown sodium thiosulphate can be calculated. Also, the unit conversion of volume must be taken care of.