Question

If 20ml of 0.5m na2so4 is mixed with 50ml of 0.2m h2so4 and 30ml of 0.4m of al2(so4)3 solution, calculate [na+], [h+], [so42-] adn [al3+].

Answer 1

Answer : The concentration of $Na^+,H^+,SO_4^{2-}\text{ and }Al^{3+}$ are 0.02, 0.02, 0.056 and 0.024 M respectively.

Solution :

First we have to calculate the moles of $Na_2SO_4,H_2SO_4\text{ and }Al_2(SO_4)_3$

$\text{Moles of }Na_2SO_4=\text{Molarity of }Na_2SO_4\times \text{Volume of }Na_2SO_4=0.5\times 0.02=0.01mole$

$\text{Moles of }H_2SO_4=\text{Molarity of }H_2SO_4\times \text{Volume of }H_2SO_4=0.2\times 0.05=0.01mole$

$\text{Moles of }Al_2(SO_4)_3=\text{Molarity of }Al_2(SO_4)_3\times \text{Volume of }Al_2(SO_4)_3=0.4\times 0.03=0.012mole$

$Na_2SO_4$ dissociates into $2Na^+$ and $SO_4^{2-}$ ions.

Concentration of $Na^+$ = $2\times 0.01=0.02M$

Concentration of $SO_4^{2-}$ = 0.01 M

$H_2SO_4$ dissociates into $2H^+$ and $SO_4^{2-}$ ions.

Concentration of $H^+$ = $2\times 0.01=0.02M$

Concentration of $SO_4^{2-}$ = 0.01 M

$Al_2(SO_4)_3$ dissociates into $2Al^{3+}$ and $3SO_4^{2-}$ ions.

Concentration of $Al^+$ = $2\times 0.012=0.024M$

Concentration of $3SO_4^{2-}$ = $3\times 0.012=0.036M$

Total concentration of $SO_4^{2-}$ = 0.01 + 0.01 + 0.036 = 0.056 M

Therefore, the concentration of $Na^+,H^+,SO_4^{2-}\text{ and }Al^{3+}$ are 0.02, 0.02, 0.056 and 0.024 M respectively.