Question

100 ml of 0.2 M H2SO4 is mixed with 400 ml of 0.05 M of Ba3(PO4)2. Calculate the concentration of Ba^2+ ions in the resulting solution. ​

Answer 1

Answer:

100 ml of 0.2 M H2SO4 is mixed with 400 ml of 0.05 M of Ba3(PO4)2. Calculate the concentration of Ba^2+ ions in the resulting solution.

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Answer 2

Answer:$Concentration\ of\ Ba^{+2}ion\ in\ resultant\ solution$$=0.08M$

Explanation:

$3H_{2}SO_{4}+Ba_{3}(PO_{4})_{2}\rightarrow 3BaSO_{4}+2H_{3}PO_{4}\\\\At\ t=0:\\Moles\ of\ H_{2}SO_{4}=0.2\times 100=20\\Moles\ of\ Ba_{3}(PO_{4})_{2}=400\times 0.05=20\\Moles\ of\ BaSO_{4}=0\\Moles\ of\ H_{3}PO_{4}=0\\\\At\ t=eqm:\\Moles of H_{2}SO_{4}=0\\Moles\ of\ Ba_{3}(PO_{4})_{2}=20-20/3=40/3\\Moles\ of\ BaSO_{4}=3(40/3)=40\\Moles\ of\ H_{3}PO_{4}=2(40/3)=80/3\\\\Concentration\ of\ Ba^{+2}ion\ in\ resultant\ solution= Moles /volume=40/400+100=40/500=0.08M$