A 20.0-mL sample of 0.300 M HBr is titrated with 0.150 M NaOH. What is the pH of the solution after 40.3 mL of NaOH have been added to the acid?
Write and balance the equation.
mols HBr = M x L = ?
mols NaOH = M x L = ?
Which reactant is in excess.
If HBr, the pH = -log(HBr)
If NaOH, the pOH = -log(NaOH) and pH + pOH = pKw = 14.
To find the pH of the solution after the addition of NaOH, we need to determine the moles of HBr before and after the reaction.
Step 1: Calculate the moles of HBr in the initial solution.
Moles of HBr = Molarity × Volume
Moles of HBr = 0.300 M × 20.0 mL = 6.00 × 10^(-3) mol
Step 2: Calculate the moles of NaOH added.
Moles of NaOH = Molarity × Volume
Moles of NaOH = 0.150 M × 40.3 mL = 6.045 ×10^(-3) mol
Step 3: Determine the limiting reactant.
HBr and NaOH react in a 1:1 molar ratio, so the limiting reactant is the one with fewer moles, which is HBr.
Step 4: Calculate the moles of HBr remaining after the reaction.
Moles of HBr remaining = Initial moles of HBr - Moles of NaOH added
Moles of HBr remaining = 6.00 × 10^(-3) mol - 6.045 ×10^(-3) mol
Moles of HBr remaining = -4.50 × 10^(-5) mol
Step 5: Calculate the final volume of the solution.
The total volume of the solution after the addition of NaOH is the sum of the initial volume of HBr and the volume of NaOH added.
Final volume = 20.0 mL + 40.3 mL = 60.3 mL
Step 6: Calculate the final concentration of HBr.
Final concentration = Moles of HBr remaining / Final volume
Final concentration = (-4.50 × 10^(-5) mol) / (60.3 mL) = -7.46 × 10^(-4) M
Note: The concentration cannot be negative, so the moles have reacted completely, and the HBr has been consumed.
Step 7: Calculate the pOH of the solution.
pOH = -log[OH-]
Since HBr and NaOH react in a 1:1 ratio, the concentration of OH- ions is equal to the concentration of NaOH.
Concentration of OH- = 0.150 M
pOH = -log(0.150) = 0.823
Step 8: Calculate the pH of the solution.
pH = 14 - pOH = 14 - 0.823 = 13.177
Therefore, the pH of the solution after 40.3 mL of NaOH have been added to the acid is approximately 13.177.
To find the pH of the solution after the titration, we need to determine the amount of excess HBr and NaOH remaining in the solution and calculate their concentrations. Then, we can use the concentrations to solve for the pH.
First, let's calculate the number of moles of HBr in the initial 20.0 mL sample:
moles HBr = volume (L) * concentration (M)
moles HBr = 0.020 L * 0.300 M
moles HBr = 0.006 moles
Since the stoichiometric ratio between HBr and NaOH is 1:1, we can say that 0.006 moles of NaOH is required to react with all the moles of HBr.
When 40.3 mL of 0.150 M NaOH is added to the acid, we can calculate the number of moles of NaOH using the same formula:
moles NaOH = volume (L) * concentration (M)
moles NaOH = 0.0403 L * 0.150 M
moles NaOH = 0.006045 moles
Now, let's find out how much HBr and NaOH are in excess:
Excess HBr = Initial moles of HBr - Moles of NaOH added
Excess HBr = 0.006 - 0.006045
Excess HBr ≈ 0 moles
(Note: Since NaOH is limiting, the excess HBr is almost negligible in this case.)
Now, let's calculate the concentration of NaOH after the addition:
Volume of NaOH after addition = Initial volume of NaOH + Volume of NaOH added
Volume of NaOH after addition = 0.020 L + 0.0403 L
Volume of NaOH after addition = 0.0603 L
Concentration of NaOH after addition:
Concentration = Moles of NaOH / Volume of NaOH after addition
Concentration = 0.006045 moles / 0.0603 L
Concentration ≈ 0.100 M
Next, we need to calculate the concentration of the resultant NaBr salt. Since the reaction between HBr and NaOH is neutralization, the moles of NaOH and HBr completely react to form NaBr salt.
Since the total volume of the solution is the initial volume of HBr (20.0 mL) plus the volume of NaOH added (40.3 mL), we can convert it to liters:
Total volume = (20.0 mL + 40.3 mL) / 1000 = 0.0603 L
Concentration of NaBr:
Concentration = Moles of NaBr / Total volume
Concentration = 0.006045 moles / 0.0603 L
Concentration ≈ 0.100 M
Now that we have the concentration of NaBr, we can find the pOH. Since NaBr is a salt of a strong base (NaOH) and a strong acid (HBr), it's the hydrolysis of water that affects the pH of the solution.
pOH = -log10[OH-]
pOH ≈ -log10[0.100 M]
pOH ≈ -log10[1.0 x 10^-1]
pOH ≈ -(-1)
pOH ≈ 1
Now, to find the pH, we can use the equation: pH + pOH = 14
pH + 1 = 14
pH ≈ 14 - 1
pH ≈ 13
Therefore, the approximate pH of the solution after 40.3 mL of NaOH has been added is around 13.