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Question: Molar conductivity of \(NH_{4}OH\) can be calculated by the equation...

Molar conductivity of NH4OHNH_{4}OH can be calculated by the equation

A

ΔºNH4OH=ΔºBa(OH)2+ΔºNH4Cl−ΔºBaCl2\Delta º_{NH_{4}OH} = \Delta º_{Ba(OH)_{2}} + \Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}}

B

ΔºNH4OH=ΔºBaCl2+ΔºNH4Cl−ΔºBa(OH)2\Delta º_{NH_{4}OH} = \Delta º_{BaCl_{2}} + \Delta º_{NH_{4}Cl} - \Delta º_{Ba(OH)_{2}}

C

ΔºNH4OH=ΔºBa(OH)2+2ΔºNH4Cl−ΔºBaCl22\Delta º_{NH_{4}OH} = \frac{\Delta º_{Ba(OH)_{2}} + 2\Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}}}{2}

D

ΔºNH4OH=ΔºNH4Cl+ΔºBa(OH)22\Delta º_{NH_{4}OH} = \frac{\Delta º_{NH_{4}Cl} + \Delta º_{Ba(OH)_{2}}}{2}

Answer

ΔºNH4OH=ΔºBa(OH)2+2ΔºNH4Cl−ΔºBaCl22\Delta º_{NH_{4}OH} = \frac{\Delta º_{Ba(OH)_{2}} + 2\Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}}}{2}

Explanation

Solution

ΔºBa(OH)2=ΔºBa2++2ΔºOH−\Delta º_{Ba(OH)_{2}} = \Delta º_{Ba^{2 +}} + 2\Delta º_{OH^{-}}

ΔºBaCl2=ΔºBa2++2ΔºCl−\Delta º_{BaCl_{2}} = \Delta º_{Ba^{2 +}} + 2\Delta º_{Cl^{-}}

ΔºNH4Cl=ΔºNH4+ΔºCl−\Delta º_{NH_{4}Cl} = \Delta º_{NH_{4}} + \Delta º_{Cl^{-}}

After substituting the above in

ΔºNH4OH=ΔºBa(OH)2+2ΔºNH4Cl−ΔºBaCl22\Delta º_{NH_{4}OH} = \frac{\Delta º_{Ba(OH)_{2}} + 2\Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}}}{2}

We get ΔºNH4OH=ΔºNH4+ΔºOH−\Delta º_{NH_{4}OH} = \Delta º_{NH_{4}} + \Delta º_{OH^{-}}