Question

the ionic strength of a 0.2M Na2HPO4 solution will be

Answer 1

Heya user this is your answer..



=1/2 [0.2x2x1*2 + 0.2x1x(-2)*2] 
=1/2 [0.4 + 0.8] 
=1/2 [1.2] 
= 0.6




Hope it helps...✌✌

Answer 2

Answer : The ionic strength of 0.2 M $Na_2HPO_4$ solution will be, 1.2 M

Solution : Given,

Concentration of $Na_2HPO_4$ solution = 0.2 M

The dissociation reaction of $Na_2HPO_4$ solution is,

$Na_2HPO_4\rightarrow 2Na^++H^++PO_4^{3-}$

Formula used for ionic strength of solution :

$\mu=\frac{1}{2}\sum C\times Z^2$

$\mu=\frac{1}{2}[2C\times (\text{Charge on }Na^+)^2+C\times (\text{Charge on }H^+)^2+C\times (\text{Charge on }PO_4^{3-})^2]$

where,

$\mu$ = ionic strength of solution

C = concentration of solution

Z = charge on ions in the solution

Now put all the given values in the above expression, we get

$\mu=\frac{1}{2}[2(0.2M)\times (+1)^2+(0.2M)\times (+1)^2+(0.2M)\times (-3)^2]$

$\mu=1.2M$

Therefore, the ionic strength of 0.2 M $Na_2HPO_4$ solution will be, 1.2 M