Dissociation constant of $NH_4OH$ is $1.8 \times 10^{-5}$. The hydrolysis constant of $NH_4Cl$ would be a) $1.8 \times 10^{-19}$ b) $5.55 \times 10^{-10}$ c) $5.55 \times 10^{-5}$ d) $1.80 \times 10^{-5}$

Dissociation constant of NH₄OH is 1.8 × 10⁻⁵. The hydrolysis constant of NH₄Cl would be a) 1.8 × 10⁻¹⁹ b) 5.55

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NH_4OH

1.8 \times 10^{-19}

5.55 \times 10^{-10}

5.55 \times 10^{-5}

1.80 \times 10^{-5}

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Step 1. $NH_4Cl$ is the salt of the strong acid HCl and the weak base $NH_4OH$ -- cationic hydrolysis, so $K_h\cdot K_b=K_w$ , giving $K_h=\dfrac{K_w}{K_b}$ . …

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Q.Dissociation constant of $NH_4OH$ is $1.8 \times 10^{-5}$ . The hydrolysis constant of $NH_4Cl$ would be

Concept understanding — Hydrolysis of Salts

Q.Dissociation constant of NH4OHNH_4OHNH4​OH is 1.8×10−51.8 \times 10^{-5}1.8×10−5. The hydrolysis constant of NH4ClNH_4ClNH4​Cl would be

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Q.Dissociation constant of $NH_4OH$ is $1.8 \times 10^{-5}$ . The hydrolysis constant of $NH_4Cl$ would be

Unlock everything free for 14 days