1. Use Kohlrausch’s law of independent ionic migration

For any strong electrolyte at infinite dilution,

$$\Lambda_m^0 = \lambda_+^0 + \lambda_-^0$$

where $\lambda_+^0$ and $\lambda_-^0$ are the limiting ionic conductivities of the cation and anion respectively.

Let

$$\lambda_{K^+}^0 = K,\quad \lambda_{Na^+}^0 = N,\quad \lambda_{Cl^-}^0 = C,\quad \lambda_{Br^-}^0 = B,\quad \lambda_{I^-}^0 = I$$

Then,

$$(\Lambda_m^0)_{KCl}=K+C,$$ $$(\Lambda_m^0)_{NaCl}=N+C,$$ $$(\Lambda_m^0)_{KBr}=K+B,$$ $$(\Lambda_m^0)_{NaBr}=N+B,$$ $$(\Lambda_m^0)_{NaI}=N+I$$

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2. Check option A

Given:

$$(\Lambda_m^0)_{KCl}-(\Lambda_m^0)_{NaCl}=(\Lambda_m^0)_{KBr}-(\Lambda_m^0)_{NaBr}$$

Substitute:

$$(K+C)-(N+C)=(K+B)-(N+B)$$ $$K-N=K-N$$

This is correct.

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3. Check option B

Given:

$$(\Lambda_m^0)_{NaBr}-(\Lambda_m^0)_{NaI}=(\Lambda_m^0)_{KBr}-(\Lambda_m^0)_{NaBr}$$

Substitute:

$$(N+B)-(N+I)=(K+B)-(N+B)$$ $$B-I=K-N$$

There is no general reason for

$$B-I=K-N$$

to be true. Hence this relation is incorrect.

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4. Check option C

Given:

$$(\Lambda_m^0)_{NaBr}-(\Lambda_m^0)_{NaCl}=(\Lambda_m^0)_{KBr}-(\Lambda_m^0)_{KCl}$$

Substitute:

$$(N+B)-(N+C)=(K+B)-(K+C)$$ $$B-C=B-C$$

This is correct.

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5. Check option D

Given:

$$(\Lambda_m^0)_{H_2O}=(\Lambda_m^0)_{HCl}+(\Lambda_m^0)_{NaOH}-(\Lambda_m^0)_{NaCl}$$

By Kohlrausch’s law:

$$(\Lambda_m^0)_{HCl}=\lambda_{H^+}^0+\lambda_{Cl^-}^0$$ $$(\Lambda_m^0)_{NaOH}=\lambda_{Na^+}^0+\lambda_{OH^-}^0$$ $$(\Lambda_m^0)_{NaCl}=\lambda_{Na^+}^0+\lambda_{Cl^-}^0$$

Therefore,

$$(\Lambda_m^0)_{HCl}+(\Lambda_m^0)_{NaOH}-(\Lambda_m^0)_{NaCl} =\lambda_{H^+}^0+\lambda_{OH^-}^0$$

which corresponds to the limiting molar conductivity associated with water via

$$H^+ + OH^- \to H_2O$$

So this relation is correct.

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6. Conclusion

The incorrect equation is:

$$\boxed{\text{B}}$$