Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added and make a graph of pH versus Va: Va = 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL.

This is the titration of a strong base with a strong acid.

Begin: 100 mL x 0.100M NaOH = 10 millimoles.
millimoles HBr added = 1.00M x mL = ?

.............NaOH + HBr ==> NaBr + H2O
initial......10mmol...0.......0......0
added.................0
change........-0............
equil.........10mmol..........

The trick here is to know what is in the solution when you are ready to determine the pH. In the case of V HBr = 0, you can see we have a solution of 10 mmoles NaOH in 100 mL = and M = mmol/mL = 10/100 = 0.1 M NaOH.
That will give you (OH^-) = 0.1 which is pOH of 1 and pH of 13.

Next point.
............NaOH + HBr ==> NaBr + H2O
initial......10.....0.......0......0
added.............1x1=1...........
change.......-1....-1.......+1.....+1
equil........9.......0.......1......1
So you have a solution of 9 mmoles NaOH and 1 mmole NaBr. NaBr is not hydrolyzed (either Na or Br) so M NaOH = mmol/mL = 9/101 = 0.0891; pOH = -log(OH^-) = 1.05 and pH + pOH = pKw = 14 so pH = 12.94 (the 101 for mL total volume is 100 to start with NaOH + 1 mL added from HBr.)

That is how you do all of the points from the beginning to the equivalence point.
The equivalence point is determined by the hydrolysis of the salt. Since it isn't hydrolyzed, the pH at the equivalence point is just the pH of pure water which is pH = 7.0.

All points past the equivalence point are done the same way with the only difference being that all of the NaOH is used and you are adding successively larger excesses of HBr.You can find which point is the equivalence point by mLNaOH x MNaOH = mLHBr x MHBr and solve for mL HBr.
Post your work if you need additional help.

1294

To find the pH at the specified volumes of acid added during the titration, we need to determine the number of moles of NaOH that react with the HBr at each volume. Then, we can calculate the concentration of the remaining OH- ions and use that information to find the pH.

Let's break down the process step-by-step:

1. Start with the initial conditions:
- Volume of NaOH solution (Va) = 100 mL
- Concentration of NaOH (Ca) = 0.100 M
- Concentration of HBr (Cb) = 1.00 M

2. Calculate the moles of NaOH present initially:
Moles of NaOH = Volume (in liters) * Concentration
Moles of NaOH = 0.100 L * 0.100 mol/L
Moles of NaOH = 0.010 mol

3. Determine the number of moles of HBr that react with NaOH at each volume (Va).
Initially, at Va = 0 mL, all the NaOH is present. So, no reaction has occurred.

For Va = 1 mL, the moles of HBr that react with NaOH can be determined using the balanced chemical equation:
NaOH + HBr → NaBr + H2O
From the balanced equation, we see that 1 mole of HBr reacts with 1 mole of NaOH.
Therefore, at Va = 1 mL, 0.01 moles of NaOH reacts with 0.01 moles of HBr.

Similarly, we can calculate the moles of HBr reacting for the other volumes using the same calculation.

4. Calculate the remaining moles of NaOH for each volume (Va).
Initial moles - moles of HBr reacted = Remaining moles of NaOH

5. Determine the concentration of OH- at each volume (Va).
Concentration of OH- = Moles of NaOH remaining / Total volume (in liters)
Total volume = Va (in liters) + 0.100 L (initial volume of NaOH)
Calculate the concentration of OH- for each volume by dividing moles by this total volume.

6. Find the pOH using the formula:
pOH = -log[OH-]

7. Calculate the pH at each volume using the equation:
pH = 14 - pOH

Once you have determined the pH at each volume, you can plot a graph of pH versus Va using the given volumes: 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL.

pH=12.94