Consider the titration of 88.0 mL of 0.150 M Ba(OH)2 by 0.600 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.
0.0 ml
12 ml
31 ml
44 ml
88 ml
You have 5 points on the titration curve. You must recognize where you are on the curve.
At zero mL, you have pure Ba(OH)2 so the pH will be determined by that concn.
At the other points, calculate how moles Ba(OH)2 you have initially, the moles HCl added. Calculate moles Ba(OH)2 remaining in solution after the neutralization of that many moles, then determine pH from that (remember it takes 2 moles HCl to equal 1 mol Ba(OH)2).
It is a good idea to determine the end point so you will know which of these points is after the end point.
Finally, if any are after the equivalence point, the pH will be determined by the amount of excess HCl present.
Post your work if you have a question about a specific point.
To calculate the pH of the resulting solution after each volume of HCl is added, we need to determine the concentration of OH- ions left in the solution. This can be done by calculating the moles of Ba(OH)2 reacted with HCl and then using the balanced equation to determine the moles of OH- ions that remain.
Here's how you can calculate the pH at each volume of HCl added:
1. Volume of HCl = 0.0 mL:
Since no HCl has been added, the concentration of OH- ions is equal to the initial concentration of Ba(OH)2. Calculate the moles of OH- ions using the formula:
Moles of OH- = Initial concentration of Ba(OH)2 * Volume of Ba(OH)2 solution (in L)
= 0.150 M * 0.088 L
= 0.0132 mol OH-
Since Ba(OH)2 dissociates into 2 OH- ions, the concentration of OH- ions is now:
Concentration of OH- = 2 * (0.0132 mol OH-) / Total volume of solution (in L)
= 2 * (0.0132 mol OH-) / 0.088 L
= 0.3 M
To calculate the pOH, use the formula:
pOH = -log10(OH- concentration)
= -log10(0.3 M)
= 0.52
Finally, to calculate the pH, subtract the pOH from 14:
pH = 14 - pOH
= 14 - 0.52
= 13.48
2. Volume of HCl = 12 mL (0.012 L):
To calculate the moles of OH- ions reacted with HCl, we need to determine the moles of HCl reacted. Since the molar ratio between Ba(OH)2 and HCl is 2:2, the moles of HCl is the same as the moles of Ba(OH)2. Calculate the moles of OH- remaining using the formula:
Moles of OH- = Initial moles of OH- ions - Moles of HCl reacted
= 0.0132 mol OH- - 0.0132 mol OH-
= 0 mol OH- ions remaining (all OH- ions have reacted)
Since there are no OH- ions remaining, the concentration of OH- ions is 0 M. Hence, the pOH is infinity (pOH = -log10(0)).
Therefore, the pH = 14 - pOH = 14 - ∞ = 14 (solution is acidic).
3. Volume of HCl = 31 mL (0.031 L):
The moles of HCl reacted is equal to the initial moles of OH- ions, as explained in the previous step.
Moles of OH- = Initial moles of OH- ions - Moles of HCl reacted
= 0.0132 mol OH- - 0.0132 mol OH-
= 0 mol OH- ions remaining
Again, there are no OH- ions, so the concentration of OH- ions is 0 M.
The pOH is infinity, so the pH = 14 - pOH = 14 - ∞ = 14 (solution is acidic).
4. Volume of HCl = 44 mL (0.044 L):
Following the same steps as above, we find that the concentration of OH- ions is 0 M, resulting in a pH of 14 (solution is acidic).
5. Volume of HCl = 88 mL (0.088 L):
The moles of OH- ions can be calculated as follows:
Moles of OH- = Initial moles of OH- ions - Moles of HCl reacted
= 0.0132 mol OH- - 0.0264 mol OH- (twice the moles of HCl reacted in the previous steps)
= -0.0132 mol OH-
Since the moles of OH- ions are negative, it means that excess HCl has been added and OH- ions have all reacted. Thus, the concentration of OH- ions is 0 M.
The pOH is infinity, so the pH = 14 - pOH = 14 - ∞ = 14 (solution is acidic).
Therefore, for all volumes of HCl added in this titration, the resulting solution is acidic with a pH of 14.