Answer to Question #96610 in General Chemistry for Brianna

Question #96610
Calculate the number of milliliters of 0.573 M Ba(OH)2 required to precipitate all of the Pb2+ ions in 177 mL of 0.403 M Pb(NO3)2 solution as Pb(OH)2. The equation for the reaction is:

Pb(NO3)2(aq) + Ba(OH)2(aq) Pb(OH)2(s) + Ba(NO3)2(aq)

___mL Ba(OH)2
1
Expert's answer
2019-10-17T07:30:53-0400
"Ba(OH)_2 + Pb(NO_3)_2 = Pb(OH)_2+Ba(NO_3)_2"

Moles of "Pb(NO_3)_2"

"n = c\\times V = 0.403 \\times 0.177 L = 0.0713 mol"


As mole ratio "n(Ba(OH)_2):n(Pb(NO_3)_2) = 1:1" , then "n(Ba(OH)_2)= 0.0713 mol"



"V=\\frac{n}{c} = \\frac{0.0713 }{0.573}=0.124 L =124 mL"


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