Question

A dilute solution of ${{H}_{2}}S{{O}_{4}}$ is made by adding $5mL$ of $3N\text{ }{{H}_{2}}S{{O}_{4}}$ to $245mL$ of water. Calculate the normality and molarity of the solution.

Answer 1

Think about the formulae that are used to calculate the molarity and the normality of any solution. Also, consider the definition of the gram equivalent weight and how you calculate the n-factor.

Complete step by step solution:
The two main formulae that we will require to calculate the normality and molarity of the final solution will be:

& {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\\ & N=n\times M \\\ \end{aligned}$$ Where, ${{N}_{1}}$ and ${{V}_{1}}$ are the normality and the volume of the original solution and ${{N}_{2}}$ and ${{V}_{2}}$ are the normality and molarity of the final solution. In the formula that relates the molarity and normality of a solution, $n$ is defined as the number of gram equivalents or the n-factor. The n-factor is defined as the number of replaceable ${{H}^{+}}$ ions that a molecule has. It can also be called the basicity of the given acid. First, let us calculate the normality of the final solution, from that we can calculate its molarity. We know the following information and we have to find the value of ${{N}_{2}}$: $$\begin{aligned} & {{N}_{1}}=3N \\\ & {{V}_{1}}=5mL \\\ & {{V}_{2}}=245mL+5mL \\\ \end{aligned}$$ So, putting these in the formula and solving for the normality of the final solution, we get: $$\begin{aligned} & 3N\times 5mL={{N}_{2}}\times 250mL \\\ & {{N}_{2}}=\dfrac{3\times 5}{250}N \\\ & {{N}_{2}}=0.06N \\\ \end{aligned}$$ **So, the normality of the final solution is $0.06N$.** Now, to find the molarity of the solution, we first have to calculate the n-factor. Let us look at the chemical formula of the given compound i.e. ${{H}_{2}}S{{O}_{4}}$. We know that sulphuric acid is a dibasic acid. This means that it has 2 replaceable protons. So, the n-factor is 2 in this case. Now, let us put the values in the formula and find the molarity of the final solution. $$\begin{aligned} & N=n\times M \\\ & 0.06N=2\times M \\\ & M=\dfrac{0.06}{2}M \\\ & M=0.03M \\\ \end{aligned}$$ **So, the molarity of the final solution is $0.03M$.** **Note:** We do not need to convert all the units to the SI units here since the millilitres are going to get cancelled anyway. Also, not that we have considered the volume of the final solution to be $250mL$ since we are adding $5mL$ of the original solution to $245mL$ of water; so, the total volume of the final solution will be $250mL$.