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

Chapter 6, Problem 96P

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QUESTION:

Aqueous sulfurous acid $$H_2SO_3$$ was made by dissolving 0.200 L of sulfur dioxide gas at $$19^{\circ}C$$ and 745 mmHg in water to yield 500.0 mL of solution. The acid solution required 10.0 mL of sodium hydroxide solution to reach the titration end point. What was the molarity of the sodium hydroxide solution?

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

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Determine the molarity of a NaOH solution through titration with sulfurous acid. Starting with the ideal gas equation we derive the concentration of a 0.200L SO?-derived H?SO? solution. Concluding with a molarity of 1.64 M for NaOH using calculated values.

QUESTION:

Aqueous sulfurous acid $$H_2SO_3$$ was made by dissolving 0.200 L of sulfur dioxide gas at $$19^{\circ}C$$ and 745 mmHg in water to yield 500.0 mL of solution. The acid solution required 10.0 mL of sodium hydroxide solution to reach the titration end point. What was the molarity of the sodium hydroxide 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 $$0.00817 \mathrm{~mol}$$