Question

The dissolution of 0.200 l of sulfur dioxide at 19 °c and 745 mmhg in water yields 500.0 ml of aqueous sulfurous acid. The solution is titrated with 13.4 ml of sodium hydroxide. What is the molarity of naoh?

Answer 1

`Molarity=1.22\ M`

Step-by-step explanation:

Given:

Pressure = 745 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 745 / 760 = 0.9803 atm

Temperature = 19 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (19 + 273.15) K = 292.15 K

Volume = 0.200 L

Using ideal gas equation as:

`PV=nRT`

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K

⇒n = 0.008174 moles

From the reaction shown below:-

`H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O`

1 mole of
`H_2SO_4` react with 2 moles of
`NaOH`

0.008174 mole of
`H_2SO_4` react with 2*0.008174 moles of
`NaOH`

Moles of
`NaOH` = 0.016348 moles

Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)

So,

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Molarity=(0.016348)/(0.0134)\ M`

`Molarity=1.22\ M`