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Step-by-Step Solution
Step 1: Identify the given quantities
• Concentration of weak acid HA, c = 0.001 \text{ mol L}^{-1} .
• Conductivity, \kappa = 2.0 \times 10^{-5} \text{ S cm}^{-1} .
• Molar conductivity at infinite dilution, \Lambda_m^\infty(\text{HA}) = 190 \text{ S cm}^2 \text{ mol}^{-1} .
Step 2: Calculate the molar conductivity of the solution, \Lambda_m
The formula for molar conductivity is
\Lambda_m = \frac{1000 \,\kappa}{c},
where c is in \text{mol L}^{-1} and \kappa is in \text{S cm}^{-1} .
Substituting the given values:
\Lambda_m = \frac{1000 \times 2.0 \times 10^{-5}}{0.001} = 20 \,\text{S cm}^2 \text{mol}^{-1}.
Step 3: Determine the degree of dissociation, \alpha
The degree of dissociation is given by
\alpha = \frac{\Lambda_m}{\Lambda_m^\infty}.
Substituting the values:
\alpha = \frac{20}{190} = \frac{2}{19}.
Step 4: Express equilibrium concentrations in terms of \alpha
For the weak acid dissociation:
\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-,
if the initial concentration is c = 0.001 \,\text{mol L}^{-1} and the degree of dissociation is \alpha , then at equilibrium:
• [\text{HA}] = c(1 - \alpha) ,
• [\text{H}^+] = c \,\alpha ,
• [\text{A}^-] = c \,\alpha .
Step 5: Use the expression for the acid dissociation constant, K_a
The acid dissociation constant is given by
K_a = \frac{[\text{H}^+] \,[\text{A}^-]}{[\text{HA}]}.
Substituting the equilibrium concentrations in terms of c and \alpha , we get
K_a = \frac{(c \,\alpha)(c \,\alpha)}{c (1 - \alpha)}
= c \,\frac{\alpha^2}{1 - \alpha}.
Here, c = 0.001 \,\text{mol L}^{-1} and \alpha = \frac{2}{19}.
Thus,
K_a = 0.001 \times \frac{\left(\frac{2}{19}\right)^2}{1 - \frac{2}{19}}.
Step 6: Numerical calculation of K_a
First compute \alpha^2 :
\alpha^2 = \left(\frac{2}{19}\right)^2 = \frac{4}{361}.
Next, compute (1 - \alpha) :
1 - \alpha = 1 - \frac{2}{19} = \frac{17}{19}.
Therefore,
\frac{\alpha^2}{1 - \alpha}
= \frac{\frac{4}{361}}{\frac{17}{19}}
= \frac{4}{361} \times \frac{19}{17}
= \frac{76}{361\,\times\,17}.
On approximate calculation, this turns out to be about 0.01238 .
Hence,
K_a = 0.001 \times 0.01238 = 1.238 \times 10^{-5}.
You can also write this as 12.38 \times 10^{-6} , which on rounding to the nearest integer for the coefficient in front of 10^{-6} is 12 \times 10^{-6} .
Step 7: Final result
Therefore, the ionization constant of the weak acid HA is
12 \times 10^{-6}.
(Rounded to the nearest integer.)
