This question has multiple parts. Work all the parts to get the most points. Cal

Question

This question has multiple parts. Work all the parts to get the most points. Calculate the pH of the solution that results when 20.0 mL of 0.1520 M lactic acid is (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) diluted to 60.0 mL with distilled water. PH = mixed with 40.0 mL of 0.0760 M NaOH solution. PH = mixed with 40.0 mL of 0.140 M NaOH solution. PH = mixed with 40.0 mL of 0.140 M sodium lactate solution. PH =

Explanation / Answer

a)

lactic acid concentration after dilution = 20 x 0.1520 / (20 +60) = 0.038 M

HCH3H5O3 ------------------------> CH3H5O3- + H+

0.038 -x x x

Ka = [CH3H5O3-][H+]/[HCH3H5O3]

1.38 x 10^-4 = x^2 / 0.038-x

x = 2.22 x 10^-3

[H+] = 2.22 x 10^-3 M

pH = -log [H+]

pH = 2.65

b)

millimoles of acid = 20 x 0.1520 = 3.04

miimoles of NaOH = 40 x 0.0760 = 3.04

it is equivalence point

salt formed = 3.04 / 60 = 0.05 M

pH = 7 + 1/2 [pKa + logC]

pH = 7 + 1/2 [3.86 + log 0.05]

pH = 8.28

c)

millimoles of NaOH = 40 x 0.140 = 5.6

NaOH remains = 5.6 - 3.04 / (20 +40) = 0.043 M

pOH = 1.37

pH + pOH = 14

pH = 12.63

d)

millimoles of lactate = 40 x 0.140 = 5.6

pH = 3.86 + log (5.6 / 3.04)

pH = 4.12

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