Question

A volume of 500.0 mL500.0 mL of 0.140 M0.140 M NaOHNaOH is added to 565 mL565 mL of 0.200 M0.200 M weak acid (?a=7.29×10−5).(Ka=7.29×10−5). What is the pHpH of the resulting buffer?

HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)

Answer 1

Moles of Naon: 0.140×500 mnd - 70 mmol Moles of acid CHA): 0.2 x 565 mm/ - 113mmy HA + Na on → Na A + H₂O moly of NaA, 70 mmol moly of HA remainin 113-70 - 43 mmol from hendersson hesselbalch equoli pH = pka + log NGA - log( 7.29x105) + log 70 4.137 + 0.211 PH= 14.35