Finding NaOH Molarity: Titration of 0.200L SO?-Derived H?SO? Solution

Here, we are going to calculate the molarity of  \(\mathrm{NaOH}\) solution.

Step 1 of 4

Calculation of the moles of \(\mathrm{SO}_{2}\) by using the ideal gas equation.

Given that,

\(\text { Volume, } V=0.2 \mathrm{~L}\)

\(\text { Temperature, } T=19{ }^{\circ} \mathrm{C}=(273.15+19) \mathrm{K}=292.15 \mathrm{~K}\)

\(\text { Pressure, } P=745 \mathrm{mmHg}=745 \mathrm{mmHg} \times \frac{1.0 \mathrm{~atm}}{760 \mathrm{~mm} \mathrm{Hg}}=0.980 \mathrm{~atm}\)

Therefore, from the ideal gas equation,

\(\begin{aligned} P V & =n R T \\ \Rightarrow n & =\frac{P V}{R T} \\ & =\frac{0.980 \mathrm{~atm} \times 0.2 \mathrm{~L}}{0.0821 \mathrm{~atm} . \mathrm{L} / \mathrm{molK} \times 292.15 \mathrm{~K}}=0.00817 \mathrm{~mol}\  \mathrm{SO}_{2} \end{aligned}\)

Thus, the moles of \(\mathrm{SO}_{2}\)  is \(0.00817 \mathrm{~mol}\)