Question

Tartaric acid, ${H_2}{C_4}{H_4}{O_6}$ is neutralized with $0.100M$ $NaOH$ . A sample of $3.0$ g of tartaric acid reacts with $45$ $mL$ of base. How concentrated is the base $?$

Answer 1

First we have to know that the concentration of a solution is the percent of the solution that is composed of the solute. The concentration of a solution is often measured in molarity ( $M$ ), which is the number of moles of solute in one litre of solution. This molar concentration ( ${c_i}$ ) is calculated by dividing the moles of solute ( ${n_i}$ ) by the total volume ( $V$ )
i.e., ${c_i} = \dfrac{{{n_i}}}{V}$ ----(1)

Complete Step By Step Answer:
The concentration of a substance is the quantity of solute present in a given quantity of solution. The SI unit for molar concentration is mol/m3.
The chemical reaction of ${H_2}{C_4}{H_4}{O_6}$ with $NaOH$ is written as follows:
${H_2}{C_4}{H_4}{O_6}\left( {aq} \right) + NaOH(aq) \to N{a_2}{C_4}{H_4}{O_6}(aq) + {H_2}O(l)$ -------(1)
To balance the equation (1), we need to multiply $NaOH$ and ${H_2}O$ by $2$ , The balance chemical reaction is:
${H_2}{C_4}{H_4}{O_6}\left( {aq} \right) + 2NaOH(aq) \to N{a_2}{C_4}{H_4}{O_6}(aq) + 2{H_2}O(l)$
We know that the atomic mass of tartaric acid is $150.087\,g/mol$ .
Then, moles of tartaric acid $= \dfrac{{3.0\;g}}{{150.087\;g/mol}} = 0.0200\,mol$
The concentration of $NaOH$ $\left( {[NaOH]} \right) = \dfrac{{moles}}{{volume}}$
$\Rightarrow [NaOH] = \dfrac{{2 \times 0.02\,mol}}{{45\,mL \times {{10}^{ - 3}}L.m{L^{ - 1}}}}$
$\Rightarrow [NaOH] = 0.888\,mol/L$ .

Note:
The percent of the solution that is composed of the solute. can be determined in one of three ways:
(1) the mass of the solute divided by the mass of the solution.
(2) the volume of the solute divided by the volume of the solution, or
(3) the mass of the solute divided by the volume of the solution.
Also, it is important to always indicate how a given percentage was calculated.