Question

How many milliliters of 0.200M NaOH solution are needed to react with 22.0 mL of a 0.490 M $NiC{{l}_{2}}$ solution?

Answer 1

Hint: First write down the balanced chemical equation for the reaction and then analyse the stoichiometry, and using it, find the volume of $NaOH$ used.

Complete step-by-step answer: In order to answer the question, we need to know about moles and molar mass. Now, matter is made up of atoms, and for the fact that matter contains mass, then the atoms should possess individual mass. Molar mass of an element or compound is the mass which houses $6\times {{10}^{23}}$ particles. For, example, the hydrogen molecule has a molar mass of $2g\,mo{{l}^{-1}}$. This means 2 grams of hydrogen contains $6\times {{10}^{23}}$atoms, and this number is also called the Avogadro’s number.
Number of moles of a compound or an element is the ratio of its given mass (the mass taken by the experiment taker) , to its molar mass. More is the number of moles, more is the concentration of the substance. Now, let us come to the question. First, we will write down the balanced chemical equation for the reaction and analyse the stoichiometry:
     \[NiC{{l}_{2}}+2NaOH\to Ni{{(OH)}_{2}}+2NaCl\]
Here, 1 mole of $NiC{{l}_{2}}$ reacts with 2 moles of sodium hydroxide to produce 1 mole of $Ni{{(OH)}_{2}}$ along with 2 moles of sodium chloride. Now, we know that the formula of molarity is $molarity=\dfrac{no\,of\,mole}{vol\,of\,solution}$, so we can calculate the number of moles of $NiC{{l}_{2}}$ as the molarity and the volume is given. So, the number of moles shall be $(22\times {{10}^{-3}}\times 0.490)=0.0108mol$. In this case, we can apply stoichiometry and find out the number of moles $NaOH$ is required is 0.0216.
In the question, we have been provided with $0.200mol\,{{L}^{-1}}$sodium hydroxide solution. So, the volume of sodium hydroxide is:
     \[\dfrac{0.0216mol}{0.200mol{{L}^{-1}}}\times 1000mL\,{{L}^{-1}}=108mL\]
So 108 mL of sodium hydroxide solution is obtained, which is the required answer to our question.

Note: The expanded formula of molarity can be written as $molarity=\dfrac{given\,mass}{molar\,mass}\times \dfrac{1000}{vol\,of\,sol(L)}$, however, the term $\dfrac{given\,mass}{molar\,mass}$ is also known as the mole fraction, hence we shorten it up.