Question 16.7: A 10.0 mL volume of 0.250 M HNO3 is titrated with 0.150 M Na......
A 10.0 mL volume of 0.250 M HNO_{3} is titrated with 0.150 M NaOH. Calculate the pH of the solution after addition of 15.0 mL of the NaOH solution.
STRATEGY
Write the neutralization reaction and set up a reaction table with amounts in moles (or millimoles). Determine if there is excess acid (before equivalence point), water (at equivalence point), or excess base (after equivalence point) once the neutralization reaction has occurred. Find the concentrations of the species remaining after the neutralization reaction by dividing the number of moles by the final volume.
| IDENTIFY | |
| Known | Unknown |
| Volume and molarity of HNO_{3} (10.0 mL, 0.250 M) | pH |
| Volume and molarity of NaOH (15.0 mL, 0.150 M ) | |
Prepare a reaction table with the number of moles of each reactant.
The number of millimoles of H_{3}O^{+} present initially is the product of the initial volume of HNO_{3} and its molarity: mmol H_{3}O^{+} initial = (10.0 mL)(0.250 mmol/mL) = 2.50 mmol. Similarly, the number of millimoles of OH^{-} added is the product of the volume of NaOH added and its molarity: mmol OH^{-} added = (15.0 mL)(0.150 mmol/mL) = 2.25 mmol. For each mmol of OH^{-} added, an equal amount of H_{3}O^{+} will disappear because of the neutralization reaction. The number of millimoles of H_{3}O^{+} remaining after neutralization is therefore
mmol H_{3}O^{+} after neutralization = mmol H_{3}O^{+} _{ initial} – mmol OH^{-} _{added}
= 2.50 mmol – 2.25 mmol = 0.25 mmol
To find the concentration H_{3}O^{+} divide the number of millimoles of H_{3}O^{+} after neutralization by the total volume (now 40.0 + 10.0 = 50.0 mL), we obtain [H_{3}O^{+}] after neutralization:
[H_{3}O^{+}] after neutralization = \frac{0.25 mmol}{50.0 mL}=5.00\times 10^{-3} M
pH = -log (5.00 × 10^{-3}) = 2.30
| Neutralization Reaction | H_{3}O^{+}(aq) | + | OH^{-}(aq) | \overset{100\%}{\longrightarrow} | 2 H_{2}O(l) |
| Before reaction (mmol) | 2.50 | 2.25 | |||
| Change (mmol) | -2.25 | -2.25 | |||
| After reaction (mmol) | 0.25 | ∼ 0 |