Question

\[\ce{\Lambda^{\circ}_m(NH_4OH)}\] [i.e., \[\ce{\Lambda^{\circ}_m(NH4OH)}\]] is equal to ______.

Options

• \[\ce{\Lambda^{\circ}_m(NH4Cl) + \Lambda^{\circ}_m(NaCl) - \Lambda^{\circ}_m(NaOH)}\]

• \[\ce{\Lambda^{\circ}_m(NaOH) + \Lambda^{\circ}_m(NaCl) - \Lambda^{\circ}_m(NH4Cl)}\]

• \[\ce{\Lambda^{\circ}_m(NH4OH) + \Lambda^{\circ}_m(NH4Cl) - \Lambda^{\circ}_m(HCl)}\]

• \[\ce{\Lambda^{\circ}_m(NH4Cl) + \Lambda^{\circ}_m(NaOH) - \Lambda^{\circ}_m(NaCl)}\]

Answer 1

\[\ce{\Lambda^{\circ}_m(NH_4OH)}\] [i.e., \[\ce{\Lambda^{\circ}_m(NH4OH)}\]] is equal to \[\ce{\mathbf{\underline{\Lambda^{\circ}_m(NH4Cl) + \Lambda^{\circ}_m(NaOH) - \Lambda^{\circ}_m(NaCl)}}}\].

Explanation:

(i) \[\ce{NH4Cl <=> NH^{+}4 + Cl-}\] 

(ii) \[\ce{NaCl <=> Na+ + Cl-}\] 

(iii) \[\ce{NaOH <=> Na+ + OH-}\]   

(iv) \[\ce{NH4OH <=> NH^{+}4 + OH-}\] 

To get equation (iv),

\[\ce{\Lambda^{\circ}_m(NH4Cl) + \ce{\Lambda^{\circ}_m(NaOH) - \ce{\Lambda^{\circ}_m(NaCl)}} = \Lambda^{0}_m(NH4OH)}\]