Question

The Ka value for HF is 3.5×10?4.

Part A

Calculate the change in pH when 2.0×10?2mol of NaOH is added to 0.50 L of a buffer solution that is 0.15 M in HF and 0.20 M in NaF.

Note: can you break it down in steps so i can understand how it was solved? Thanks

Answer 1

The pH of the buffer solution can be determined by Henderson - Hasselbal ch equati on which is, pH-pk. +log conjugate base] acid] F (NaF)] HF -log (35x10-+)+log 10[0.20 M 0.15 M 3.5808

The buffer solution contains 0.15 M HF and 0.20M in NaF Consi der that the volume of the buffer solution does not change on the addition of NaOH. Hence the specified volume, 0.5 L,is the volume of the resultant solution. The moles of HF concentration of HF × volume of the resultant solution = 0.1514 × 0.5L 0.075 mol The moles of NaF concentration of Nar × volume of the resultant solution 0.20 14 × 0.5 L -0.1 mol The moles of NaOH addedis 2.0x10mol The weak aci d of the buffer consumes all the strong base. Hence, the amounts changed by each compound changedis written below the chemical equati on HF + NaOH -2.0×10-2 mol (0 mol) +NaF 0.075 mol-2.0×10-2 mol (0.055 mol) 0.1 mol +2.0×10-2 mol (0.12 mol) After consumption of the added bas e: The pH of the buffer solution can be determined by Henderson Hasselbalch equation which is pHplog conjugate base pH = pK, + log acid] F (NaF)] HF 0.12mol 0.5L -log (3.5x10-+)+1og 0055 mol 0.5L = 3.79 Us the change in the pH- 3.79 -3.58 0.21