Calculate the number of milliliters of 0.689 M Ba(OH)2 required to precipitate all of the Pb2+ ions in 123 mL of 0.663 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

118ml of 0.689M Ba(OH)₂

Explanation:

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

moles Ba(OH)₂ needed = moles of Pb(NO₃)₂ reacted

moles = Molarity x Volume

(Molarity x Volume)Ba(OH)₂ = (Molarity x Volume)Pb(NO₃)₂

0.689M Ba(OH)₂ · V[Ba(OH)₂] = 0.663M Pb(NO₃)₂ · 123-ml

V[Ba(OH)₂] needed = 0.663M · 123-ml / 0.689M = 118.4-ml ≈ 118-ml (3 sig.figs.)

We need 118 mL of Ba(OH)2

Explanation:

Step 1: Data given

Molarity of Ba(OH)2 = 0.689 M

Volume of a 0.663 M Pb(NO3)2 solution = 123 mL = 0.123 L

Step 2: The balanced equation

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

Step 3:

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of Ba(OH)2 = 1

⇒with Ca = the concentration of Pb(NO3)2 = 0.663 M

⇒with Va = the volume of Pb(NO3)2 = 0.123 L

⇒with a = the coefficient of Pb(NO3)2 = 1

⇒with Cb = the concentration of Ba(OH)2 = 0.689 M

⇒with Vb = the volume of Ba(OH)2 = TO BE DETERMINED

0.663M * 0.123L = 0.689 * Vb

Vb = 0.118 L = 118 mL

We need 118 mL of Ba(OH)2

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa