Consider the titration of 50.00 ml .250 M HBr with .100 M Ba(OH)2.
What is the pH after the following additions of Ba(OH)2 to the 50 ml of hydrobromic acid?
0 ml. 15 ml. 25 ml. 50ml. 51ml.
Five separate problems can take a long time. Exactly what do you understand (and not understand)?
Mostly just setting up my reaction to use for each problem
To determine the pH at each addition of Ba(OH)2, we need to understand the reaction that takes place between HBr and Ba(OH)2. This is an acid-base reaction where HBr acts as the acid and Ba(OH)2 acts as the base. The reaction equation can be written as:
HBr + Ba(OH)2 -> BaBr2 + 2H2O
Since HBr is a strong acid and Ba(OH)2 is a strong base, this reaction will go to completion.
Now, let's calculate the moles of HBr initially present in the 50.00 ml of 0.250 M HBr solution:
moles of HBr = volume (in liters) x molarity
= 0.050 L x 0.250 mol/L
= 0.0125 mol HBr
Since the reaction is stoichiometric, each mole of HBr reacts with one mole of Ba(OH)2 to form BaBr2. Therefore, the moles of Ba(OH)2 required to react with all the HBr is also 0.0125 mol.
Now, let's analyze the pH at each addition of Ba(OH)2:
1. 0 ml Ba(OH)2: No reaction has occurred yet, so the pH is determined solely by the HBr. Since HBr is a strong acid, it completely dissociates in water to give H+ ions. Therefore, the pH is determined by the concentration of H+ ions, which can be calculated using the initial concentration of HBr: pH = -log[H+].
[H+] = Molarity of HBr (since it is a strong acid)
= 0.250 M
pH = -log(0.250) ≈ 0.602
2. 15 ml Ba(OH)2: By adding 15 ml of Ba(OH)2, we have exceeded the required amount to neutralize all the HBr. The excess Ba(OH)2 will react to form a basic solution. The moles of Ba(OH)2 added is given by:
moles of Ba(OH)2 = volume (in liters) x molarity
= 0.015 L x 0.100 mol/L
= 0.0015 mol Ba(OH)2
Since 1 mole of Ba(OH)2 reacts with 2 moles of HBr, only half of the moles of Ba(OH)2 will react to form BaBr2, leaving the other half as excess hydroxide ions (OH-).
Remaining moles of Ba(OH)2 = 0.0125 mol - (0.0015 mol / 2) = 0.0125 mol - 0.00075 mol = 0.01175 mol
The volume of the final solution after adding 15 ml of Ba(OH)2 is 50 ml + 15 ml = 65 ml.
Now, to calculate the pH, we need to find the concentration of OH- ions from the unreacted excess Ba(OH)2. The concentration can be calculated as:
OH- concentration = moles of OH- / volume (in liters)
= 0.00075 mol / 0.065 L
= 0.0115 M
Since OH- ions are a strong base, we can calculate the pOH from the concentration of OH- ions:
pOH = -log(0.0115) ≈ 1.94
Now, to find the pH, we can subtract the pOH value from 14 (14 is the sum of pH and pOH):
pH = 14 - pOH
= 14 - 1.94
= 12.06
Therefore, the pH after adding 15 ml of Ba(OH)2 is approximately 12.06.
Similarly, you can repeat this process to find the pH after adding 25 ml, 50 ml, and 51 ml of Ba(OH)2.