Question

A solution of NaOH was standardized by gravimetric titration of a known quantity of the primary standard, potassium hydrogen phthalate:

The was then used to find the concentration of an unknown solution of${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$

${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{+}\mathbf{2}\mathbf{NaOH}\mathbf{\to }{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

A 10.00-mL aliquot of ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$solution required 57.911 g of NaOH solution to reach the phenolphthalein end point. Find the molarity of ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$

Short answer

The molarity of sulfuric acid in the titration of $\mathrm{NaOHVs}{\mathrm{H}}_2{\mathrm{SO}}_4$is 0.305M.

Step-by-step solution

Define the molarity

The total number of moles of solute per liter of solution is known as a solution's molarity. Because, unlike mass, the volume of a system fluctuates with changes in physical conditions, the molality of a solution is affected by changes in physical parameters of the system such as pressure and temperature. Molarity is represented by the letter M, which stands for a molar.

The molarity of a solution is defined as one gramme of solute dissolved in one litre of solution.

Step 2: Calculate the molarity of H2SO4

The molarity of a solution can be given by the formula

$Molarity(M)=\frac{(Molesofsolute(inmol)}{(Volumeofsolution(inL)}$

The molarity of sulfuric acid in the titration of$\mathrm{NaOHVs}{\mathrm{H}}_2{\mathrm{SO}}_4$

$\mathrm{molNaOH}=(0.057911\mathrm{kg}\mathrm{solution})(\mathrm{O}.10531\mathrm{mol}/\mathrm{kg}\mathrm{solution})$

$=6.0986\mathrm{mmol}$

The ratio of sodium hydroxide to sulfuric acid is 2 : 1

$[H_{2}S{O}_4]=\frac{0.5\times 6.098mmol}{10.00mL}$

$=0.305\mathrm{M}$

Thus, the molarity of sulfuric acid in the titration of$\mathrm{NaOH}\mathrm{Vs}{\mathrm{H}}_2{\mathrm{SO}}_4$ is 0.305M