Question 17.7: Weak Acid–Strong Base Titration pH Curve A 40.0-mL sample o...

(a) The equivalence point occurs when the amount (in moles) of added base equals the amount (in moles) of acid initially in the solution. Begin by calculating the amount (in moles) of acid initially in the solution.
The amount (in moles) of KOH that must be added is equal to the amount of the weak acid.

mol \ HNO_2 = 0.0400 \ \cancel{L} \times\frac{0.100 \ mol}{\cancel{L}}

= 4.00 × 10^{-3} mol

mol KOH required = 4.00 × 10^{-3} mol

Calculate the volume of KOH required from the number of moles of KOH and the molarity

volume KOH solution = 4.00 × 10^{-3} \ \cancel{mol}×\frac{1 \ L}{0.200 \ \cancel{mol}}

= 0.0200 L KOH solution
= 20.0 mL KOH solution

(b) Use the concentration of the KOH solution to calculate the amount (in moles) of OH^- in 5.00 mL of the solution.

= 1.00 \times 10^{-3} mol OH^-

Prepare a table showing the amounts of HNO_2 and NO_2^- before and after the addition of 5.00 mL KOH. The addition of the KOH stoichiometrically reduces the concentration of HNO_2 and increases the concentration of NO_2^-.

Since the solution now contains significant amounts of a weak acid and its conjugate base, use the Henderson– Hasselbalch equation and pK_a for HNO_2 (which is 3.34) to calculate the pH of the solution.

pH = pK_a + \log\frac{[base]}{[acid]}

= 3.34 + \log\frac{1.00 \times 10^{-3}}{3.00 \times10^{-3}}

= 3.34 – 0.48 = 2.86

(c) At one-half the equivalence point, the amount of added base is exactly half the initial amount of acid. The base converts exactly half of the HNO_2 into NO_2^-, resulting in equal amounts of the weak acid and its conjugate base. The pH is therefore equal to pK_a.

pH = pK_a + \log\frac{[base]}{[acid]}

= 3.34 + \log\frac{2.00 \times 10^{-3}}{2.00 \times10^{-3}}

= 3.34 + 0 = 3.34