50.0mL of 0.01M Ba(OH)2 is added to 150.0mL of 0.01M HNO3. What is the new pH of the solution?

50 mL x 0.01M Ba(OH)2 = 0.5 millimoles.

150 mL x 0.01 M HNO3 = 1.5 mmols.
....Ba(OH)2 + 2HNO3 ==> Ba(NO3)2 + 2H2O
I...0..........1.50........0.........0
added 0.5...............................
C....-0.5......-1.00......0.5........
E.......0.......0.5.........0.5........
(HNO3) = 0.5 millimols/200 mL = 0.0025 M
I work in millimoles when there are so many zeros. For example, for the Ba(OH)2 part.
mols = M x L = 0.01 x 0.050 = 0.0005 mol.
But millimols = 50 x 0.01 = 0.500.
You can change to mols in the above if you wish by dividing millimoles by 1000.

To determine the new pH of the solution after mixing Ba(OH)2 and HNO3, we need to first identify the reaction that takes place between the two compounds. Ba(OH)2 is a strong base, and HNO3 is a strong acid. When a strong base reacts with a strong acid, a neutralization reaction occurs, resulting in the formation of a salt and water.

The balanced chemical equation for the reaction between Ba(OH)2 and HNO3 is:

Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

From the equation, we see that 1 mole of Ba(OH)2 reacts with 2 moles of HNO3 to produce 1 mole of Ba(NO3)2 and 2 moles of water. Since the concentrations of Ba(OH)2 and HNO3 are both 0.01M, we can calculate the number of moles present.

Using the formula for moles, moles = concentration * volume (in liters), we have:

Moles of Ba(OH)2 = 0.01M x (50.0mL/1000mL/L) = 0.0005 moles
Moles of HNO3 = 0.01M x (150.0mL/1000mL/L) = 0.0015 moles

Now, let's determine the limiting reactant to find out which compound will be completely consumed. The stoichiometric ratio between Ba(OH)2 and HNO3 is 1:2. Since Ba(OH)2 has fewer moles than HNO3, it is the limiting reactant. Thus, all of the Ba(OH)2 will be consumed to form Ba(NO3)2.

Now, let's calculate the new concentration of HNO3 (and Ba(NO3)2) in the solution. Initially, we had 0.0015 moles of HNO3 in 150.0mL of solution. However, after the reaction, we will have the same number of moles but in a new total volume. The final volume of the solution is the sum of the initial volumes of Ba(OH)2 and HNO3:

Final volume = 50.0mL + 150.0mL = 200.0mL

Now, we can calculate the new concentration of HNO3:

New concentration of HNO3 = Moles of HNO3 / Final volume (in liters)
= 0.0015 moles / (200.0mL/1000mL/L)
= 0.0075 M

Since the reaction between Ba(OH)2 and HNO3 is a neutralization reaction, the resulting solution will be neutral with a pH of 7.