a. The stoichiometry equation is HNO2(aq) + KOH(aq) → KNO2(aq) + H2O(l). b. The volume required to reach the equivalence point is 20.0 mL. c. The pH after adding 5.00 mL of KOH is 6.7.
a. The stoichiometry equation for this titration is:
HNO2(aq) + KOH(aq) → KNO2(aq) + H2O(l)
b. To calculate the volume required to reach the equivalence point, we need to use the stoichiometry of the reaction.
The mole ratio of HNO2 to KOH is 1:1, so the number of moles of KOH required is equal to the number of moles of HNO2:
0.100 M HNO2 × 0.0400 L = 0.00400 mol HNO2
Since the mole ratio is 1:1, we need 0.00400 mol KOH. Using the concentration and the number of moles, we can calculate the volume of KOH:
0.00400 mol KOH ÷ 0.200 M KOH = 0.0200 L = 20.0 mL
c. The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution.
However, for this question, we need to calculate the pH after adding 5.00 mL of KOH.
This will go beyond the buffer region, so we need to consider the excess OH- ions.
The pH is calculated using the concentration of OH- ions and the concentration of HNO2:
OH-: [KOH] × (volume added/Vol. original + volume added) = 0.200 M × (5.00 mL/40.0 mL + 5.00 mL) = 0.025 M
HNO2: 0.100 M
Now, we can use the equation pH = pKa + log([A-]/[HA]) to calculate the pH:
pH = 7.2 + log(0.025 / 0.100) = 6.7