Question

4. Merri Thon and Rundy Miles are both runners on the ARC track team (don'tlyou 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

Answer 1

Answer:-

Given:-

molar concentration of lactic acid (HLac) i.e [HLac] = 0.4250 M

pK_a of lactic acid (HLac) = 3.85

pH of solution of lactic acid (HLac) = ?

As we know that

pK_a of acid = - log[equilibrium constant (K_a) of acid]

So

pK_a = - log(K_a)

therefore

pK_a of lactic acid = - log(equilibrium constant (K_a) of lactic acid )

3.85 = - log(equilibrium constant (K_a) of lactic acid )

log(equilibrium constant (K_a) of lactic acid ) = - 3.85

On taking antilog of the above value

equilibrium constant (K_a) of lactic acid =  0.000141

Also we know that

HLac $\rightleftharpoons$ H⁺ + Lac^-

Inital 0.4250 M 0 0

Change -x    +x +x

Equilibrium (0.4250 - x)    +x +x

So

equilibrium constant (K_a) of lactic acid = [H⁺][ Lac^- ] / [HLac]

equilibrium constant (K_a) of lactic acid = x $\times$ x / (0.4250 - x)

equilibrium constant (K_a) of lactic acid = x² / (0.4250 - x)

Suppose in denominator x << 0.4250 M then it can be neglected in above equation

equilibrium constant (K_a) of lactic acid = x² / 0.4250

0.000141 = x² / 0.4250

x² = 0.000141 $\times$ 0.4250

x² = 0.000059925

x² = 0.59925 $\times$ 10^-4

x = $\sqrt{}$ 0.59925 $\times$ 10^-4

x = 0.774$\times$ 10^-2

Therefore

molar concentration of H⁺ ion i.e [H⁺] = x = 0.774$\times$ 10^-2 M

molar concentration of Lac^- ion i.e [Lac^-] = x = 0.774$\times$ 10^-2 M

Since we know that

pH = - log[H⁺]

pH = - log[0.774$\times$ 10^-2]

pH = - [ log(0.774) + ( - 2log10) ]

pH = - [ (-0.1112) + ( - 2) ]

pH = - (-0.1112 - 2)

pH = - (-2.1112 )

pH = 2.1 (i.e the answer)