Question

van't hoff factor for an infinitely dilute solution of NaHSO4 is

Answer 1

Vant Hoff factor i for an infinitely dilute solution of NaHSO4 is (1) 1/2 (2) 1/3 (3) 2 (4) 3

Answer 2

Answer:

The van't hoff factor (i) after infinite dilution of solution of NaHSO₄ will be equal to 3.

Explanation:

We know that NaHSO₄ is a weak acid. Therefore it's dissociation occurs in equilibrium.

          $NaHSO_{4}$   ⇄   $Na^{+}+HSO^{-}_{4}$

HSO₄⁻ further dissociates:

          $HSO^{-}_{4}$    ⇄    $H^{+}+SO^{2-}_{4}$

Total number of ions  = 3   $(Na^{+},H^{+},SO^{2-}_{4})$

To find the van't hoff factor (i)

             $i=1+(n-1)\alpha$

where α is degree of ionization and $\alpha =1$ for infinitely dilution

 ∴              $i=1+(3-1)(1)\\ i=3$

Therefore  for an infinitely dilute solution of NaHSO₄ van't hoff factor equals to three.