Question

A 0,205 M NaOH solution is used to titrate $20.0\mathrm{mL}$of a solution of ${\mathrm{H}}_2{\mathrm{SO}}_4$. If $45.6\mathrm{ml}$, of the $\mathrm{NaOH}$solution is required to reach the endpoint, what is the molarity of the ${\mathrm{H}}_2{\mathrm{SO}}_4$solution? $(0,7)$

$\left.{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{NaOH}\left(\alpha q\right)⟶2{\mathrm{H}}_2\mathrm{OND}\right)+{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\alpha q\right)$

Short answer

Molarity${\mathrm{H}}_2{\mathrm{SO}}_4=\frac{2\times 0.00234{\mathrm{moleH}}_2{\mathrm{SO}}_4}{0.02\mathrm{L}}$

$=0.234{\mathrm{MH}}_2{\mathrm{SO}}_4$

Step-by-step solution

Given information

That are given numbers are calculated by : (1)

Metric Molarity Mole divide by

Explanation

Step 1 Given $45.6\mathrm{ml}$of$0.205\mathrm{MNaOH};20.0{\mathrm{mlH}}_2{\mathrm{SO}}_4=0.02{\mathrm{LH}}_2{\mathrm{SO}}_4$.

Need molarity of ${\mathrm{H}}_2{\mathrm{SO}}_4$

Step 2 Plan

Metric Molarity Mole divide by

$45.6\mathrm{mlL}$moles $\mathrm{NaOH}$moles${\mathrm{H}}_2{\mathrm{SO}}_4{\mathrm{MH}}_2{\mathrm{SO}}_4$

Factor factor factor liters

Step 3 Equalities /conversion factors

$$\begin{gathered} 1L\mathrm{NaOH}=0.205\mathrm{moleNaOH} \\ \frac{1L}{0.205\mathrm{mole}}=\frac{0.205\mathrm{mole}}{1L} \\ 1L\mathrm{NaOH}=1000\mathrm{mlNaOH} \\ \frac{1L}{1000\mathrm{ml}}=\frac{1000\mathrm{ml}}{1L} \end{gathered}$$

Given information 

Equation are:

Explanation 

2 mole $\mathrm{NaOH}=$l mole${\mathrm{H}}_2{\mathrm{SO}}_4$

$\frac{2moleNaOH}{1mole{H}_2{\mathrm{SO}}_4}$ and $\frac{\text{Imole}{\mathrm{H}}_2{\mathrm{SO}}_4}{2\text{mole}\mathrm{NaOH}}$

Step 4 Set up problem

$$\begin{gathered} 45.6\mathrm{mlSeOH}\times \frac{1\mathrm{LNeOH}}{1000\mathrm{mlNaOH}}\times \frac{0.205\mathrm{moleNeOH}}{1L\mathrm{SaeOH}}\times \frac{\text{lmole}{\mathrm{H}}_2{\mathrm{SO}}_4}{2\mathrm{mole}\text{,}\mathrm{SeOH}} \\ =0.00234{\mathrm{moleH}}_2{\mathrm{SO}}_4 \\ \text{Molarity}{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{2\times 0.00234{\mathrm{mole}}_2{\mathrm{SO}}_4}{0.02\mathrm{L}} \\ =0.234{\mathrm{MH}}_2{\mathrm{SO}}_4 \end{gathered}$$