electrochemistry

Question: ionisation constant of a weak acid ($HA$) in terms of $\Lambda _{m}^{\infty }$ and ${{\Lambda }_{m}}$ is

1) ${{K}_{a}}=\frac{C\Lambda _{m}^{\infty }}{({{\Lambda }_{m}}-{{\Lambda }^{\infty }})}$

2) ${{K}_{a}}=\frac{C\Lambda _{m}^{2}}{\Lambda _{m}^{\infty }(\Lambda _{m}^{\infty }-{{\Lambda }_{m}})}$

3) ${{K}_{a}}=\frac{C{{(\Lambda _{m}^{\infty })}^{2}}}{\Lambda _{m}^{\infty }(\Lambda _{m}^{\infty }-{{\Lambda }_{m}})}$

4) None of these

Solution:

Explanation: No Explanation

Conductors and Conductance

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)