Question

A 1.000 L sample of HCl gas at 25∘C and 732.0 mm Hg was absorbed completely in an aqueous solution that contained 6.948 g of Na2CO3 and 246.0 g of water. A.What is the pH of the solution? B.What is the freezing point of the solution? C.What is the vapor pressure of the solution? (The vapor pressure of pure water at 25 ∘C is 23.76 mm Hg.)

Answer 1

First we will calculate the density of HCl gas at 25 ⁰ c, and 732 mm of hg,

then PV=nRT, n= molar mass =mass of gas/mol wt=m/M

SUBSTITUTING THIS FOR n,

PV=m/M*RT

P/RT=m/M*V

WE KNOW DENSITY =D=m/V

REARRANGING,

P/RT=D/M

D=MP/RT,

D=36.5*0.963/0.0821*298 ( Pressure conversion mm of hg atm & temp to K)

D=35.149/24.43

=1.438 g/l

but the volume of water is 246 gm=246 ml

1000 ml-246 ml=754 ml volume of HCl gas, we can calculate the density to 754 ml by cross multiplying,

1.438=1000 then

? =754,

=1.084 g/l

then from we can calculate the concentration ,

=1.084/36.5=0.029 for 1000 ml

for 754 ml= it will concntrated to 0.04

therefore concentration HLC =[0.04]

THEN WE WILL CLACULATE THE

NA2CO3 CONC.

WT =6.948 IN 246 ML WATER,MOLWT =105.98

105.98=1000 ML=1 MOLAR,

6.948=1000 ML =?=0.065 FOR 100 ML THEN FOR 246 ML HOW MANY?

IT WILL CONCENTRATE TO =1.04 ,

Na2co3=[1.04], by knowing conc. we can calculate the ph of solution

c) it can be calculated using Raoults low by knowing the

fractions of the solution

P_solution = P°_A χ_A + P°_B χ_B