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Find the saponification value of an oil sample 1.5g refluxed with 25 ml of 0.5 N KOH required 15 ml of 0.5 N HCI for the residual titration. The blank titration reading was 25 ml of 0.5 N HCI.
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written 3.4 years ago by
teamques10
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Given:
Oil sample $=1.5g$
Normality of $\text{KOH}=0.5N$
Normality of $\text{HCl}=0.5N$
Blank tiration $=25ml$
Formula:
$\text{Saponification value}(mg/g)=[\text{(Blank level)-(Titration Volume)}]\times \text{Normality of KOH}\times \text{Equivalent weight of KOH}$
Solution:
$\begin{align*} \text{Saponification value}(mg/g) &=\dfrac{[(25ml)-(15ml)]\times 0.5N\times 56}{1.5}\\[2ex] &=\dfrac{10\times 0.5\times 56}{1.5}\\[2ex] &=\dfrac{280}{1.5}\\[2ex] &=186.67\end{align*}$
Therefore, Saponification value is 186.67
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