Question 17.6: Strong Base–Strong Acid Titration pH Curve A 50.0-mL sample ...
Strong Base–Strong Acid Titration pH Curve
A 50.0-mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid. Calculate the pH:
(a) after adding 30.00 mL of HNO_3
(b) at the equivalence point
(a) Begin by calculating the initial amount of NaOH (in moles) from the volume and molarity of the NaOH solution. Because NaOH is a strong base, it dissociates completely, so the amount of OH^- is equal to the amount of NaOH.
moles NaOH =0.0500\cancel{L}\times \frac{0.200 \ mol}{1 \ \cancel{L}}
= 0.0100 mol
moles OH^- = 0.0100 \ mol
Calculate the amount of HNO_3 (in moles) added at 30.0 mL from the molarity of the HNO_3 solution.
moles HNO_3 added = 0.0300 \ L \times \frac{0.200 \ mol}{1 \ L}
= 0.00600 mol HNO_3
As HNO_3 is added to the solution, it neutralizes some of the OH^-. Calculate the number of moles of OH^- remaining by setting up a table based on the neutralization reaction that shows the amount of OH^- before the addition, the amount of H_3O^+ added, and the amounts left after the addition.
| OH^−(aq) + H_3O^+(aq)\longrightarrow 2 \ H_2O(l) | ||||
| Before addition | 0.0100 mol | ≈0.00 mol | ||
| Addition | — | 0.00600 mol | ||
| After addition | 0.0040 mol | 0.00 mol | ||
Calculate the OH^- concentration by dividing the amount of OH^- remaining by the total volume (initial volume plus added volume).
Calculate the pOH from [OH^-].
[OH^-]=\frac{0.0040 \ mol}{0.0500 \ L + 0.0300 \ L}
= 0.0500 M
pOH = -\log(0.0500)
=1.30
Calculate the pH from the pOH using the equation
pH + pOH = 14.
pH = 14 – pOH
= 14 – 1.30
= 12.70
(b) At the equivalence point, the strong base has completely neutralized the strong acid. The [H_3O^+] at 25 °C from the ionization of water is 1.00 \times 10^{-7} M and the pH is therefore 7.00.
pH = 7.00