Question

0.106 g of $N{{a}_{2}}C{{O}_{3}}$ neutralizes 40.0 mL of ${{H}_{2}}S{{O}_{4}}$.Hence, normality of ${{H}_{2}}S{{O}_{4}}$ solution is (write 0.06 as 6 in the answer):

Answer 1

When sodium carbonate ($N{{a}_{2}}C{{O}_{3}}$) is treated with sulfuric acid , neutralization reaction occurs whereby, an $O{{H}^{-}}$ from the base combines with an ${{H}^{+}}$ from the acid to form ${{H}_{2}}O$ and a salt $N{{a}_{2}}S{{O}_{4}}$ (sodium sulfate) and $C{{O}_{2}}$ which is been liberated.

Complete step by step solution:
-We are asked to find the normality of the solution which is formed by adding 0.106 g of $N{{a}_{2}}C{{O}_{3}}$ to 40.0 mL of ${{H}_{2}}S{{O}_{4}}$. As we know, in chemistry normality is one of the expressions which is used to measure the concentration of a solution.
- Normality is often referred to as the equivalent concentration of a solution and is abbreviated as ‘N’. It is commonly used as a measure of reactive species in a solution and during titration reactions and in situations which involve acid-base chemistry. It can be mathematically represented as follows
$Normality=\dfrac{Gram\text{ }equivalent\text{ }of\text{ }solute~}{Volume\text{ }of\text{ }solution\text{ }in\text{ }litre}$\
- As we know, the gram equivalent weight is calculated by dividing the molecular weight of solute by the number of equivalents per mole of solute. Also, the number of moles present in a sample can be obtained by dividing the given mass of the sample by its molar mass. Therefore the number of moles present in 0.106 g of $N{{a}_{2}}C{{O}_{3}}$ can be given as below (molar mass of sodium carbonate is 106)
$0.106\text{ }g\text{ N}{{\text{a}}_{2}}C{{O}_{3}}=\dfrac{0.106}{106}mol=2\times {{10}^{-3}}\text{ }equivalents$
Since 40 ml of ${{H}_{2}}S{{O}_{4}}$ is being neutralized, the equivalents of ${{H}_{2}}S{{O}_{4}}$ can be written as follows
$Equivalents\text{ }of\text{ }{{H}_{2}}S{{O}_{4}}=\dfrac{40N}{1000}equivalents$
Where N is the normality. Since both the reactants are neutralized in the solution, they both have same number of equivalents and the above two quantities can be equated as follows
$\dfrac{40N}{1000}=2\times {{10}^{-3}}$
On solving the above equation we get the value of N as follows
$N=0.05$

Thus the normality of the solution is 0.05 N. Since in question it’s given as to write 0.06 as 6 in the answer, the final answer will be 5 N.

Note: Do not confuse between the terms molarity and normality. The term molarity is expressed as the number of moles of solute per liter of solution while normality is a measure of concentration that is equal to the gram equivalent weight per liter of solution