Question

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

Answer 1

$$Ba(OH)_2 + Pb(NO_3)_2 = Pb(OH)_2+Ba(NO_3)_2$$

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 $n(Ba(OH)_2)= 0.0713 mol$

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