234 e-exmotic pressure and (2) donie strengtb of a solution containin 0.15 gmol

Question

234

e-exmotic pressure and (2) donie strengtb of a solution containin 0.15 gmol and cancentration 7189tor bove NaCI solution together with NasSO in 0.01 M 2 kp Which salt category does sodium acetate belong to? (a) salts of strong acids and strong acetate. G) Caleulafe the bydrolytie -1.01x10 I. Calculate ) the (s lution density 1002 kuru', R-82.0578 atmecmrmok-a)- 5 strength what is te i nie is the ionic ? (18%) Na CHu sals of weak acids and strong bases,(e) salts of strong acids and wealk bases, (d) salts the appropriate of weak acids and weak bases. 2) Write down the appropriate hydrolytic reaction of sodium constant K of sodium acetate at 25°C, and/K, = 1.75x10(4) Find the degree of hydrolyiyhora. 0.02 molar solution of sodium acetate at 25°C (12%) Reactats A and B are placed in a reaction vessd at zere time, where Cas Cao. The following reactions occur at constant volume berr M is the desired product. If both reactions are edivity ef Mwith respect to Ri seceed

Explanation / Answer

1) Osmotic pressure ()= imRT-----------(1)

Where, m is the molality of solution

R=82.058 atm-cm3/gmol-K=0.082058 atm-m3/gmol-K

T= Temperature in Kelvin=25+273.15=298.15K

i=number of ions in which solute dissociates

Mass of NaCl=0.15g mol

Mol wt of NaCl=58.44g/mol

Mass of water=1000g or 1Kg

Density of solution= 1002kg/m3=1.002g/cm3

Molality (m)=moles of solute/Kg of solvent= 0.15

                                                                       58.44×1Kg of water =0.0025m

i=2 since NaCl dissociates into two ions

Place all values in (1)

= 2×0.0025×0.082058×298.15=0.122 atm

2) Ionic strength= 1/2cizi2

c is the Molar concentration

z is the charge on ion

mass of solution=0.15g of NaCl+1000g of water=1000.15g

density= 1.002g/cm3

volume=1000.15/1.002=998.15cm3=0.998L

Molarity= 0.15/58.44×0.998= 0.0025 Molar (same as water as density of solution is 1.002g/cm3)

NaClè Na++ Cl-

Ionic strength=0.5 ((0.0025×12)+(0.0025×12))=0.5(0.0025+0.0025)=0.5×0.005=0.025M

3) Ionic strength of NaCl with 0.01M Na2SO4

Na2SO4 ---->2Na++SO42-

NaClè Na++ Cl-

It dissociates into 2Na+ ions and 1SO42- ions

NaCl dissociates in Na+ and Cl-

Ionic strength= 0.5 ((0.0025×12)+(0.0025×12)+(0.01×2×12)+(0.01×(-2)2))

= 0.5 (0.0025+0.0025+0.02+0.04)= 0.0325M

2) CH3COONa is salt of weak acid strong base

Hydrolytic reaction:

CH3COO- +H2O------->CH3COOH + OH-

Kh= Kw/Ka

Kh= hydrolysis constant

Kw= ionic product of water= 1.01×10-14

Ka= acid dissociation constant=1.75×10-5

Kh= 1.01×10-14/1.75×10-5= 0.577×10-9

Degree of hydrolysis

h= CKh=CKw/Ka

C=0.02M

Kh= 0.577×10-9

h= 0.02×0.577×10-9=0.2×0.577×10-10= 0.339×10-5

           

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