Question

NaCall, BH=85 Nay + H₂O = Hut Nacht Nazo-1M, PH=9) Naz+ H₂O Hat Nach Hx=14-7-log o.lx10 6 -121-pka tog Oslo Lys og Ocelote/5=14-pka-logo.com =7-logo Pha= 14-16-log BS 3=14-pha-loals I Pka=145-4 Ipka = 14-te-573 5 pha=5] Phat 4. Merri Thon and Rundy Miles are both runners on the ARC track team (don't you just love their names....). After a very strenuous practice of spirited repeat 800's, their muscles are sore and tired. In Rundy's physiology course, he has just learned that this soreness may be caused by lactic acid, HLac, which builds up in the body during anaerobic activities. Help Rundy out by...... Calculating the pH of a 0.4250 M solution of lactic acid. PK, = 3.85 Ich a (cre spre tial in na gen lished xyge juilib n

Answer 1

Lactic acid ( CH₃CH(OH)CO₂H ) is a weak acid. It dissociate as

HC₃H₅O₃ + H₂O -----> H₃O⁺ + C₃H₅O₃^-

pH = −log[H₃O⁺]

The [H₃O⁺] can be calculated using a reaction table

H3O+ Hoi +90 +% HGH,O₃ H₂O Initial 0.4250 large change - x negligible final 10.4250-1 Large

The equilibrium constant Ka is given by:
Ka =[?_??⁺] [?_??_??_?^-]/[??_??_??_?]
    = x²/ (0.4250 - x)
Given pK_a = -log K_a = 3.85

therefore K_a = 10^-3.85 = 0.0001 and is very small and 0.4250-x ~0.4250

therefore we can equate x² = 0.4250 $\times$ 10^-3.85 = 0.00004250

                                     ^$\therefore$    x = 6.5 $\times$ 10^-3 M = [?_??⁺]

Hence pH = -log [?_??⁺] = -log 6.5 $\times$ 10^-3 = -(-2.18) = 2.18