Final answer:
After adding 30.0 mL of 0.175 M NaOH to 30.00 mL of 0.125 M HCOOH, we are past the equivalence point. The excess OH- ions from NaOH produce a pH of 12.40.
Step-by-step explanation:
To calculate the pH after 30.0 mL of 0.175 M NaOH has been added to 30.00 mL of 0.125 M HCOOH, we first need to determine if we've reached the equivalence point. The equivalence point is reached when the moles of NaOH added equals the moles of HCOOH present.
Moles of HCOOH = Volume of HCOOH (L) × Molarity of HCOOH = 0.03000 L × 0.125 M = 0.00375 mol
Moles of NaOH = Volume of NaOH (L) × Molarity of NaOH = 0.03000 L × 0.175 M = 0.00525 mol
Since we have added more moles of NaOH (0.00525 mol) than there are moles of HCOOH (0.00375 mol), we are past the equivalence point. The excess moles of NaOH will determine the pH of the solution.
Excess moles of NaOH = 0.00525 mol - 0.00375 mol = 0.00150 mol
Now we calculate the concentration of OH- ions:
Volume of the mixture = Volume of HCOOH + Volume of NaOH = 0.03000 L + 0.03000 L = 0.06000 L
[OH-] = Excess moles of NaOH / Volume of the mixture = 0.00150 mol / 0.06000 L = 0.025 M
We then calculate the pOH and subsequently the pH:
pOH = -log(0.025) = 1.60
pH = 14.00 - pOH = 14.00 - 1.60 = 12.40