Question

Aqueous sulfurous acid (H2SO3) was made by dissolving 0.200 L of sulfur dioxide gas at 19°C and 745 mmHg in water to yield 500.0 mL of solution. The acid solution required 16.9 mL of sodium hydroxide solution to reach the titration end point. What was the molarity of the sodium hydroxide solution?

Answer 1

Answer:

The molarity of the sodium hydroxide solution is 0.97 M

Explanation:

Step 1: Balanced reaction of H2SO3

SO2 + H2O → H2SO3

Step 2= Data given

Volume of SO2 = 0.2L

Temperature of SO2  = 19°C = 292.15 Kelvin

Pressure = 745 mmHg = 0.980263 atm

Via the ideal gas law we have this formula:

PV = nRT

with P = pressure of the gas = 0.980263 atm

with V = volume of the gas = 0.2L

with n = number of moles = TO BE DETERMINED

with R = Gas constant =  0.082057 L atm K−1 mol−1

Step 3: Calculating number of moles of SO2

0.980263 * 0.2 = n * 0.082057 * 292.15

n = 0.0082 moles

Step 4: Calculating number of moles of H2SO3

Since the ratio of the balanced equation is 1:1 for SO2 and H2SO3 this means that there is also 0.0082 moles of H2SO3

Step 5: Calculating moles of NaOH

The second reaction  is H2SO3 + 2 NaOH → Na2SO3 + 2H2O

This means for 1 mole of H2SO3 consumed there is needed 2 moles of NaOH

So, for 0.0082 moles of H2SO3 there is needed 0.0164 moles of NaOH

Step 6: Calculating molarity of NaOH

The molarity of the NaOH solution can be calculated as:

Molarity = Moles / volume

Molarity = 0.0164 moles / 0.0169 L

Molarity = 0.97 M

The molarity of the sodium hydroxide solution is 0.97 M