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Question

0.01 moles of a weak acid HA(Ka=2.0x10-6) is dissolved in 1.0Lof0.1MHCl solution. The degree of dissociation of HA is_______ x10-5 (Round off to the Nearest Integer). Assume the degree of dissociation <<1


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Solution

Step 1: Dissociation of weak and strong acid:

  • The weak acid will dissociate as

HAH++A-

  • The strong acid will dissociate as

HClH++Cl-

The actual formula is Ka=H+A-HA

Step 2: Finding the concentration terms:

C=nVV=1LofHClso,C=n

so, HClH++Cl-0.010.010.01

So, the concentration of reactant and product at initial and at equilibrium concentration.

HAH++A-C0.010.10Ceq0.01(1-α)0.1+0.01α0.01α0.010.1

Step 3: Finding the dissociation constant value

On substituting the values and on rearranging the equation we get,

Ka=H+A-HA=[0.1][0.01α]0.012x10-6=0.1xα2x10-6=10-1xαα=2x10-610-1α=2x10-5

Hence, the value of the degree of dissociation of HA is 2×10-5.


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