Question

Calculate the equivalent weight of $KH{C_2}{O_4}$​ (potassium acid oxalate) when it is neutralized by $KOH$ . Give an answer in the form of $\dfrac{{Eq.Wt}}{{64}} = x$ i.e. find the value of $x$.

Answer 1

Equivalent weight of any element can be found by taking the ratio of their molecular weight and dividing by the change in their oxidation number. In this question, we first need to find the change in the oxidation number of the oxalate ion that is, ${C_2}{O_4}^{ - 2}$ ion. The reaction of neutralization is:
$KH{C_2}{O_4} + KOH \to {K_2}{C_2}{O_4} + {H_2}O$

Formula used: $\dfrac{{molecular .wt}}{n} = equivalent\;{{ }}weight$
Where $n$is the number of change in oxidation number or the number of replaceable ${H^ + }$ ions.

Complete step by step answer:
it is first required to find the equivalent weight of the compound in question that is potassium acid oxalate. For this as mentioned, we must first write down the reaction of neutralization that the compound is currently going through. The reaction takes place in the following path:
$KH{C_2}{O_4} + KOH \to {K_2}{C_2}{O_4} + {H_2}O$
This means that in the compound potassium oxalate will react with potassium hydroxide and the hydrogen gets replaced by a potassium ion. This means that the number of replaceable ions is only $1$ . The molecular weight of the compound in the question is $128$ . Now if these values along with the others are plugged into the above equation we get,
$\dfrac{{128}}{1} = equivalent\;{{ }}weight$
$equivalent{{ }}weight = 128$
Therefore, now that we have found the equivalent weight, we can use that to find the unknown variable $x$.
If we plug in the value of the equivalent weight into the reaction that is given in the question that is, $\dfrac{{Eq.Wt}}{{64}} = x$ we will get the answer in the following steps,
$\dfrac{{Eq.Wt}}{{64}} = x$
$\dfrac{{128}}{{64}} = x$
$2 = x$

Therefore, we now know that the value of $x$ is $2$.

Note: Equivalent weight can be defined as the weight of a compound that can displace one gram of hydrogen, eight grams of oxygen and $35.5$ grams of chlorine.
It is important to write the reaction that is mentioned in questions that talk about neutralization. This is done because we have to find the number of ${H^ +}$or $O {H^ -}$ that are being removed or replaced from the compound.
The value of $n$ mentioned can also represent the change in the oxidation number that a particular element in the compound goes through.