Question

Equivalent conductance of saturated baso4 is 400and specific conductance is 8×1/100000.the ksp of baso4 is?

Answer 1

we can calculate solubility from that specific conductance x 1000/ equivalent conductance

Answer 2

The solubility product of Barium sulphate is $4\quad \times \quad 10^{ -8 }$

Explanation:

At first, we have to calculate the solubility of a $B a S O_{4}$ salt.

Formula:

$Solubility\quad =\quad \frac { k\quad \times \quad 1000 }{ \Lambda  }$

Where,

$k\quad =\quad { specific\quad conductance\quad }=\quad 8\quad \times \quad 10^{ -5 }\quad { Ohm }^{ -1 }{ cm }^{ -1 }$

$\Lambda \quad =\quad { equivalent\quad conductance\quad}=\quad 400\quad { Ohm }^{ -1 }{ cm }^{ 2 }eq^{ -1 }$

Now put all the given values in the above formula, we get the solubility.

$Solubility\quad =\quad \frac { 8\quad \times \quad 10^{ -5 }\quad \times \quad 1000 }{ 400 }$

$Solubility\quad =\quad 2\quad \times \quad 10^{ -4 }$

Now, we have to calculate the value of $K_{sp}$ of $BaSO_{4}$

$BaSO_{ 4 }\quad \rightarrow \quad Ba^{ 2+ }\quad +\quad SO_{ 4 }^{ 2- }$

The equilibrium reaction will be,

$k_{ sp }\quad =\quad \left[ Ba^{ 2+ } \right] \left[ SO_{ 4 }^{ 2- } \right]$

$k_{ sp }\quad =\quad s\quad \times \quad s$

$k_{ sp }\quad =\quad s^{ 2 }\quad ({ where }\quad 's'\quad { is\quad the\quad solubility) }$

$k_{ sp }\quad =\quad \left( 2\quad \times \quad 10^{ -4 } \right) ^{ 2 }$

$k_{ sp }\quad =\quad 4\quad \times \quad 10^{ -8 }$