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