Question

$$A^{\circ}$$ for $$NaCl$$, $$HCl$$ and $$NaAc$$ are 126.4,425.9 and 91.0 S $$cm^2 mol^{-1}$$ respectively, Calculate $$A^{\circ}$$ for $$HAc$$

Answer 1

Given,

$$\wedge _{NaCl}^0= \wedge _{Na^+}^0+ \wedge _{Cl^-}^0$$=126.4....(1)

$$\wedge _{HCl}^0= \wedge _{H^+}^0+ \wedge _{Cl^-}^0$$=425.9....(2)

$$\wedge _{NaHc}^0= \wedge _{Na^+}^0+ \wedge _{Ac^-}^0$$=425.9....(3)

Adding (ii) and (iii) and then subtracting (i)

$$\wedge _{HAc}^0=\wedge _{H^+}^0+ \wedge _{Ac^-}^0=425.9 +91.0-126.4$$

$$ 390.5 \, S \, cm^2\, mol ^{-1}$$