A 20.0 mL sample of 0.687 M HBr is titrated with a 0.216 M NaOH solution. The pH after the addition of 10.0 mL of NaOH is __________.
A. 0.413
B. 0.386
C. 0.163
D. 0.579
E. 0.237
To determine the pH after the addition of 10.0 mL of NaOH, we need to calculate the concentration of HBr remaining and then find the corresponding pH.
Step 1: Calculate the number of moles of HBr initially present.
Moles of HBr = concentration of HBr × volume of HBr solution
Moles of HBr = (0.687 M) × (20.0 mL/1000 mL/1 L) = 0.01374 moles
Step 2: Calculate the number of moles of NaOH added.
Moles of NaOH = concentration of NaOH × volume of NaOH solution
Moles of NaOH = (0.216 M) × (10.0 mL/1000 mL/1 L) = 0.00216 moles
Step 3: Determine the limiting reactant.
The limiting reactant is the one with fewer moles. In this case, NaOH is the limiting reactant because it has fewer moles (0.00216 moles) compared to HBr (0.01374 moles).
Step 4: Calculate the number of moles of HBr remaining.
Moles of HBr remaining = initial moles of HBr - moles of NaOH added
Moles of HBr remaining = 0.01374 moles - 0.00216 moles = 0.01158 moles
Step 5: Calculate the concentration of HBr remaining.
Concentration of HBr remaining = moles of HBr remaining / volume of solution
Concentration of HBr remaining = (0.01158 moles) / (30.0 mL /1000 mL / 1 L) = 0.386 M
Step 6: Calculate the pOH.
pOH = -log10 (concentration of HBr remaining) = -log10 (0.386) = 0.413
Step 7: Calculate the pH.
pH + pOH = 14
pH = 14 - pOH = 14 - 0.413 = 13.587
Therefore, the pH after the addition of 10.0 mL of NaOH is approximately 13.587.
The correct answer is not provided in the options.
To find the pH after the addition of NaOH, we need to consider the reaction that occurs between HBr and NaOH. HBr is a strong acid, while NaOH is a strong base. When they react, they undergo a neutralization reaction to form water and a salt.
The balanced chemical equation for the reaction between HBr and NaOH is:
HBr + NaOH → NaBr + H₂O
From the balanced equation, we can see that 1 mole of HBr reacts with 1 mole of NaOH, and they produce 1 mole of water. This means that the moles of HBr will be equal to the moles of NaOH at the equivalence point.
First, let's find the moles of HBr initially present in the 20.0 mL sample:
moles of HBr = volume (L) × concentration (M)
= 20.0 mL × (1 L / 1000 mL) × 0.687 M
= 0.01374 mol
Next, let's find the moles of NaOH added:
moles of NaOH = volume (L) × concentration (M)
= 10.0 mL × (1 L / 1000 mL) × 0.216 M
= 0.00216 mol
Since the reaction is a 1:1 stoichiometry, the moles of HBr remaining after the reaction will be:
moles of HBr remaining = moles of HBr initially - moles of NaOH added
= 0.01374 mol - 0.00216 mol
= 0.01158 mol
Finally, we can find the concentration of HBr after the addition of NaOH:
final concentration of HBr = moles of HBr remaining / volume (L)
= 0.01158 mol / (20.0 mL × (1 L / 1000 mL))
= 0.579 M
Now, to find the pH of the solution after the addition of NaOH, we need to recall that HBr is a strong acid. Strong acids dissociate completely in water. Therefore, the concentration of HBr is equal to the concentration of H+ ions in solution.
pH = -log[H+]
= -log(0.579)
= 0.237
Therefore, the pH after the addition of 10.0 mL of NaOH is approximately 0.237.
The correct answer is E. 0.237.