Q. 10.95 - An Introduction to General, Organic, and Biological Chemistry

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? $(0,7)$

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

Step-by-Step Solution

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 1: Given information

That are given numbers are calculated  by : (1)

Metric Molarity Mole divide by

Step 2: 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

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

Step 3: Given information

Equation are: 

Step 4: 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

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