Question

Suppose you're doing a research project with some thermophilic bacteria, which can live and even thrive in very hot, acidic environments. (Think "Yellowstone hotsprings.") You place a culture of these bacteria in 250.0 mL of a buffer solution containing 0.30 M formic acid (HCOOH, K_a = 1.8 ✕ 10^-4) and 0.20 M sodium formate (HCOONa) and let them stew at 80 degrees Celsius just before leaving the lab for lunch. While you're gone, a labmate, jealous of your good looks or your technical prowess or both and knowing this particular bacterium will die at a pH above 4.50, adds a scoop (about 2.75 g) of NaOH to your thermophilic brew. Will your bacteria survive? Figure it out by calculating the pH of the mixture.

Please show your work if possible! Thanks!

Answer 1

initially:

mmol of formic acid = MV = 0.3*250 = 75

mmol of formate = MV = 0.2*250 = 50

then, after addition of

m = 2.75 g of NaOH

mol of NaOH = mass/MW = 2.75/40 = 0.06875 mol --> 0.06875*10^3 = 68.75 mmol

then, there is reaction

formate increases, formic acid decreases

mmol of formic acid = = 75-68.75 = 6.25

mmol of formate = 50+68.75 = 118.75

pH = pKa + log(formate/formic acid)

pH = 3.75 + log(118.75/6.25)

pH = 5.02875

since pH > 4.5, they will not survive