Answer: 33 mL of 0.4 M Ba(OH2) must be added.
Step-by-step explanation:
We will solve this problem by tracking the number of moles in the solutions.
Consider the balanced chemical equation for the dissolution of Ba(OH)2:
Ba(OH)2 (aq) --> Ba^2+ (aq) + 2 OH- (aq)
From the coefficients we see that 2 moles of OH- ions are produced for every 1 mole of Ba(OH)2.
For NaOH:
NaOH (aq) --> Na^+ (aq) + OH- (aq),
so 1 mole of OH- ions are produced for every 1 mole of NaOH.
Number of OH- ions
From the information we gathered:
0.4 mol/L Ba(OH2) x 2 mol OH- / (1 mol Ba(OH)2) =0.8 mol/L OH-
0.3 mol/L NaOH x 1 mol OH- / (1 mol NaOH) = 0.3 mol/L OH-
We are asked to find the volume, X, required to give a solution of 0.5 M in OH-.
(0.8 mol/L of OH-) x (X) = 0.8X mol OH-
(0.3 mol/L of OH-) x (0.050 L) = 0.015 mol OH-
Our goal is:
(0.5 mol/L of OH-) x (X + 0.050 L) = 0.5(X + 0.050) mol OH-
Setting the number of moles equal:
0.8X + 0.015 = 0.5(X + 0.050)
0.8X + 0.015 = 0.5X + 0.025
0.3X = 0.010
X = 0.033 L
In other words, the volume of 0.4 M Ba(OH)2 required is 33 mL.