Excess Na_2SO_4(aq) is added to a 2.45 times 10^2 mL sample of industrial waste

Question

Excess Na_2SO_4(aq) is added to a 2.45 times 10^2 mL sample of industrial waste containing Ba^2+ ions. If 25.5 g of BaSO_4(s) are precipitated from the reaction, what was the molar concentration of Ba^2+ in the original sample? Ba^2+(aq)+Na2SO_4(aq) rightarrow BaSO_4(s)+ 2Na^+(aq) M Excess Na_2SO_4(aq) is added to a 2.45 times 10^2 mL sample of industrial waste containing Ba^2+ ions. If 25.5 g of BaSO_4(s) are precipitated from the reaction, what was the molar concentration of Ba^2+ in the original sample?

Explanation / Answer

Na2SO4 V = 245 mL

Ba+2 = ?

m = 25.5 g of BaSO4

preicpitate, find Ba+2

so..

mol of BaSO4 = mass/MW = 25.5/233.43 = 0.1092404 mol of BaSO4

ratio is 1:1 so 1 mol of Ba+2 precipitates BaSO4

0.1092404 mol of Ba+2 = 0.1092404 mol of BaSO4

so

[Ba+2] = mol/V = 0.1092404 / (245*10^-3) = 0.44587 M

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