Question

A. What is the pH of 2L of a buffer solution containing 80.0 grams of lactic acid (CH3CH(OH)COOH) (90.08 g/mol) and 100.0 grams of sodium lactate (CH3CH(OH)COONa) (112.06 g/mol)?

B. What will be the pH of the solution formed from adding 20mL of 8.0 M HCl to the buffer described in Part A of this question? (Ignore the change in volume)

For part A, because we aren't given pKa, am I right to assume that we'd be finding the molarity of the acid and then using pH = -log[H3O] to find the pH? Is it okay to ignore the sodium lactate or am I thinking about this in the wrong way?

Thanks for your help!

Answer 1

We should find ent the mine of pkg of test x exid from other sunrises (text boxes or google search) for lasts acid pka is 3.86 mass of [sodium lactate] = mass of sodium lactate_g)/meter mass of volume (in glmul)(in l) = 100.0g/112.06 glmul times 2l = 0/4462 [acid] almostequalto [sodium lactate] ph = pka = 3.86 20 ml 80m hcl = 20ml/1000ml/l times 8.0 mol /l = 0.16 moles [sodium lactate] = 2.6 times 0.446x mole/l - 0.16 mul /2l = 0.3662 m [acid] = 2l times u.444 muil + 0.16 moles/2l = 0.524m [sodium lactate]/ [acid] = log 0.6988 = -0.1557 ph = pka + log [sodium lactate]/[acid] 3.86 - 0.1557 = 3.70