Chemistry, asked by bhadsonsaini7744, 1 year ago

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+].

Answers

Answered by BarrettArcher
38

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.

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