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

Answer:

$Molarity=1.22\ M$

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,

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.016348}{0.0134}\ M$

$Molarity=1.22\ M$