To calculate the pH of each scenario, we'll need to consider the reactions that take place when the acidic solution (HCOOH) reacts with the basic solutions (NaOH and sodium formate). We'll calculate the pH for each case separately.
The reaction of HCOOH (formic acid) with NaOH (sodium hydroxide) results in the formation of sodium formate and water:
HCOOH + NaOH → HCOONa + H2O
The reaction of HCOOH with sodium formate (HCOONa) results in a buffer solution containing the conjugate acid-base pair, HCOOH and HCOONa:
HCOOH + HCOONa ⇌ HCOOH + HCOO-
Let's go through each scenario:
B. Mixing HCOOH with 25.0 mL of NaOH solution (0.160 mol/L):
First, calculate the moles of each reactant:
Moles of HCOOH = volume (L) × concentration (mol/L) = 0.020 L × 0.200 mol/L = 0.004 mol
Moles of NaOH = 0.025 L × 0.160 mol/L = 0.004 mol
Since the moles of HCOOH and NaOH are equal, they will react completely, and no excess of either will be left after the reaction. This is a neutralization reaction. The resulting solution will contain sodium formate (NaHCOO) and water (H2O). The moles of NaHCOO will be 0.004 mol.
Calculate the concentration of sodium formate:
Volume of resulting solution = volume of HCOOH + volume of NaOH = 0.020 L + 0.025 L = 0.045 L
Concentration of sodium formate = moles / volume = 0.004 mol / 0.045 L ≈ 0.0889 mol/L
Now, use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:
pH = pKa + log([A-]/[HA])
Given: pKa = -log(Ka) = -log(1.8 × 10^-4) ≈ 3.74
[A-] = concentration of HCOO- = 0.0889 mol/L
[HA] = concentration of HCOOH = 0.200 mol/L
pH = 3.74 + log(0.0889/0.200) ≈ 3.38
C. Mixing HCOOH with 25.0 mL of NaOH solution (0.200 mol/L):
Follow the same steps as in scenario B, but with the concentration of NaOH as 0.200 mol/L.
Moles of NaOH = 0.025 L × 0.200 mol/L = 0.005 mol
Concentration of sodium formate = 0.005 mol / 0.045 L ≈ 0.1111 mol/L
pH = 3.74 + log(0.1111/0.200) ≈ 3.33
D. Mixing HCOOH with 25.0 mL of sodium formate solution (0.200 mol/L):
The resulting solution will be a buffer solution containing HCOOH and HCOONa. Calculate the moles of sodium formate:
Moles of sodium formate = 0.025 L × 0.200 mol/L = 0.005 mol
Concentration of sodium formate = 0.005 mol / 0.045 L ≈ 0.1111 mol/L
Use the Henderson-Hasselbalch equation:
pH = 3.74 + log(0.1111/0.200) ≈ 3.33
So, for scenarios B and C, the pH of the resulting solutions is approximately 3.33, and for scenario D, the pH is approximately 3.33 as well.