Solveeit Logo
Login

Question

Physics Question on Electric Current

Specific conductance of o.1 ,M ,HA is 3⋅75×10−4 ohm−1 cm−13 \cdot 75 \times 10^{-4} \,ohm ^{-1} \, cm ^{-1}. Find the dissociation constant KaK _{ a } of HAHA, if λ∞(HA)=250 ohm−1 cm2 mol−1:\lambda_{\infty}( HA )=250 \, ohm ^{-1} \, cm ^{2} \,mol ^{-1}:

A

1.0×10−51.0 \times 10^{-5}

B

2.25×10−42.25 \times 10^{-4}

C

2.25×10−52.25 \times 10^{-5}

D

2.25×10−132.25 \times 10^{-13}

Answer

2.25×10−52.25 \times 10^{-5}

Explanation

Solution

λm=1000k0.1=1000×3.75×10−40.1=3.75\lambda_{m}=\frac{1000 k}{0.1}=\frac{1000 \times 3.75 \times 10^{-4}}{0.1}=3.75
α=λmλm∞=3.75250=1.5×10−2\alpha=\frac{\lambda_{m}}{\lambda_{m}^{\infty}}=\frac{3.75}{250}=1.5 \times 10^{-2}
Ka=Cα2=0.1×(1.5×10−2)2K_{a}=C \alpha^{2}=0.1 \times\left(1.5 \times 10^{-2}\right)^{2}
=2.25×10−5=2.25 \times 10^{-5}