Discuss the ionisation of weak electrolytes (Ostwald’s dilutio

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 Multiple Choice QuestionsShort Answer Type

191.

What are electrolytes and non-electrolytes?

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 Multiple Choice QuestionsLong Answer Type

192. Discuss different types of electrolytes.
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193. What is a degree of ionisation ?
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194.

Discuss the ionisation of weak electrolytes (Ostwald’s dilution law).


As the weak electrolytes are ionised to a small extent, there is a state of equilibrium between the unionised electrolyte and the ions formed in solution. To illustrate.this, let us consider a weak electrolyte HA to be dissolved in water, a be the degree of ionisation and C be the molar concentration of HA.
                                  HA     H+(aq) + A-(aq)
Initial  conc.                C            0               0
Equilibrium conc.                         
Applying law of chemical equilibrium,

                 

where Ka is ionisation (or dissociation) constant of the weak acid HA.



This equation is often referred to as Ostwald’ dilution law equation since the degree of dissociation of a weak electrolyte is very small as compared to unity.



Thus, the degree of ionisation of a weak electrolyte is inversely proportional to the square root of the molar concentration of the solution of the electrolyte.

Hence the degree of ionisation of weak electrolyte goes on increasing with the decrease in molar concentration i.e. with dilution and it reaches the maximum value (unity) in very dilute solution. Thus, all weak electrolytes undergo almost complete ionisation at infinite dilution.
For a weak base BOH:
Let us consider a weak base BOH dissolved in water, a be the degree of ionisation and C be the molar concentration of BOH.

                         
Initial conc.           C              0                 0
Equilibrium                                  
conc. 
Applying law of chemical equilibrium,



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 Multiple Choice QuestionsShort Answer Type

195.

Write the correct balanced net ionic equation for the reaction whose equilibrium constant at 298K is:
(i) Ka(C6H5COOH) = 6·3 × 10–5
(ii) Ka(H2C2O4) = 5·4 × 10–2
(iii) Ka ( HSO3) = 2·8 × 10–7
(iv) Kb(OCl) = 9·1 × 10–7

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196. Write correctly the balanced net ionic equation for the reaction whose equilibrium constant at 298K is:

(i) Kb(CH3NH2) = 4·4 × 10–5
(ii) KaS) = 1·0 × 10–7
(iii) Ka (HCN) = 4·0 × 10–10
(iv) Kb(NH3) = 1·8 × 10–5
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197. Calculate the degree of ionisation and hydronium ion concentration of a 0·02 M acetic acid at 298K. The dissociation constant of acid is 1·8 × 10–5 at 298K.
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 Multiple Choice QuestionsLong Answer Type

198. At 298K, 0·01M solution of acetic acid is 1·34% ionised. What is the ionisation constant Ka for acetic acid?
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199. Discuss in brief Arrhenius concept of acids and bases.
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 Multiple Choice QuestionsShort Answer Type

200.

What is the fate of proton in aqueous medium?

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