Molar conductivity of a weak acid HQ of conce | vedclass

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(A) For a weak acid, $K_a = C\alpha^2$ and $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.$K_a = C \left(\frac{\Lambda_m}{\Lambda_m^0}\right)^2$$pK_a = -\lo

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) For a weak acid, $K_a = C\alpha^2$ and $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.$K_a = C \left(\frac{\Lambda_m}{\Lambda_m^0}\right)^2$$pK_a = -\log K_a = -\log C - 2\log \Lambda_m + 2\log \Lambda_m^0$$pK_a(HQ) - pK_a(HZ) = \log \left(\frac{C_{HZ}}{C_{HQ}}\right) + 2\log \left(\frac{\Lambda_{m(HZ)}}{\Lambda_{m(HQ)}}\right)$Given $\frac{\Lambda_{m(HQ)}}{\Lambda_{m(HZ)}} = \frac{1}{30}$, so $\frac{\Lambda_{m(HZ)}}{\Lambda_{m(HQ)}} = 30$.Since $\lambda_{Q^{-}}^0 = \lambda_{Z^{-}}^0$ and $\lambda_{H^+}^0$ is the same for both, $\Lambda_{m(HQ)}^0 = \Lambda_{m(HZ)}^0$.$\Delta pK_a = \log \left(\frac{0.02}{0.18}\right) + 2\log(30)$$\Delta pK_a = \log \left(\frac{1}{9}\right) + 2\log(30) = -2\log 3 + 2(\log 3 + \log 10) = -2\log 3 + 2\log 3 + 2 = 2$.

Similar Questions

$\Lambda _m^\circ$ for $NaCl$,$HCl$ and $NaAc$ are $126.4$,$425.9$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda _m^\circ$ for $HAc$.

The limiting molar conductivities of the given electrolytes at $298 \ K$ follow the order: $\lambda^{\circ}_{(K^{+})} = 73.5, \lambda^{\circ}_{(Cl^{-})} = 76.3, \lambda^{\circ}_{(Ca^{2+})} = 119.0, \lambda^{\circ}_{(SO_{4}^{2-})} = 160.0 \ S \ cm^{2} \ mol^{-1}$.

Which one of the following graphs between molar conductivity $(\Lambda_m)$ versus $\sqrt{C}$ is correct?

The resistance of a $0.2 \ M$ solution of an electrolyte is $30 \ \Omega$ and conductivity is $1.2 \ S \ m^{-1}$. What is the value of the cell constant (in $cm^{-1}$)?

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