Calculate the pHpH of an aqueous solution containing the following: (a) 0.80M0.80M lactic acid and 0.40M0.40M lactate ion (b) 0.30MNH30.30MNH 3​ and 1.50MNH4+1.50MNH 4+​

To calculate the pH of the solutions, we will use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the base (or the conjugate base) and [HA] is the concentration of the acid (or the conjugate acid).

(a) For the lactic acid/lactate ion solution, we first need to know the pKa of lactic acid. The pKa of lactic acid is 3.86. Plugging the given concentrations into the Henderson-Hasselbalch equation gives us:

pH = 3.86 + log(0.40/0.80) pH = 3.86 + log(0.5) pH = 3.86 - 0.301 pH = 3.56

(b) For the NH3/NH4+ solution, we need to know the pKa of NH4+. The pKa of NH4+ is 9.25. However, since we are dealing with a base and its conjugate acid, we need to use the pOH form of the Henderson-Hasselbalch equation, which is pOH = pKb + log([A-]/[HA]). The pKb is the base dissociation constant and can be found from the pKa by the equation pKb = 14 - pKa. So, the pKb of NH3 is 14 - 9.25 = 4.75. Plugging the given concentrations into the Henderson-Hasselbalch equation gives us:

pOH = 4.75 + log(0.30/1.50) pOH = 4.75 + log(0.2) pOH = 4.75 - 0.699 pOH = 4.05

To find the pH, we subtract the pOH from 14 (since pH + pOH = 14), so the pH of the solution is 14 - 4.05 = 9.95.