Answer:
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.)
Answer:
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
