What is the ph of a solution made by combining 157 ml of 0.35 m nac2h3o2 with 139 ml of 0.46 m hc2h3o2? the ka of acetic acid is 1.75 × 10-5?

The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar

Then, we calculate the pH as follows:
pKa of acetic acid = -log(1.75 × 10^-5) = 4.7569
pH = pKa +  log ([salt] / [acid]) 
     = 4.7569 + log(0.1856 / 0.216)
     = 4.691