Question

250 ml of 0.10 M K2SO4 solution is mixed with 250
0.20 M KCl solution. The concentration of K + ions in
resulting solution will be :
(1) 0.1M
(2) 0.4M
(3) 0.2M
(4) 0.8M
​

Answer 1

Answer : The correct option is, (3) 0.2 M

Explanation :

First we have to calculate the moles of $K_2SO_4$ and $KCl$.

$\text{Moles of }K_2SO_4=\text{Concentration of }K_2SO_4\times \text{Volume of solution in liters}=0.10mole/L\times 0.25L=0.025moles$

$\text{Moles of }KCl=\text{Concentration of }KCl\times \text{Volume of solution in liters}=0.20mole/L\times 0.25L=0.05moles$

Now we have to calculate the moles of $K^+$.

As we know that, 1 mole of $K_2SO_4$ dissociates to give 2 mole of $K^+$ ion and 1 mole of $SO_4^{2-}$ ion.

So, the mole of $K^+$ ion = $2\times 0.025mole=0.05mole$

and,

As we know that, 1 mole of $KCl$ dissociates to give 1 mole of $K^+$ ion and 1 mole of $Cl^-$ ion.

So, the mole of $K^+$ ion = 0.05 mole

Total number of moles of $K^+$ ion = 0.05 + 0.05 = 0.1 mole

Now we have to calculate the concentration of $K^+$ ion.

$\text{Concentration of }K^+ion=\frac{\text{Moles of }K^+ion}{\text{Volume of solution in liters}}=\frac{0.1mole}{0.5L}=0.2mole/L=0.2M$

Therefore, the concentration of $K^+$ ion is, 0.2 M