Calculate the pH of the solution that results when 25.0 mL of 0.1650 M lactic ac
ID: 488008 • Letter: C
Question
Calculate the pH of the solution that results when 25.0 mL of 0.1650 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 50.0 mL with distilled water. pH = mixed with 25.0 mL of 0.165 M NaOH solution. pH = mixed with 25.0 mL of 0.230 M NaOH solution. pH = mixed with 25.0 mL of 0.230 M sodium lactate solution. pH =Explanation / Answer
(a) molarity of acid in diluted solution = 0.1650/2 = 0.0825 M
HLa <==> H+ + La^-
ka =1.38*10^-4
1.38*10^-4 = x^2/0.0825-x
x = 0.0033
[H+] = 0.0033M
pH = -log[h+] = 2.48
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(b)
moles of Lactic acid present = 0.025 L * 0.165 M = 4.123*10^-3 moles
Moles of NaOH added = 0.025 L *0.165 M = 4.123*10^-3 moles
At this point all the lactic acid has converted to lactate.
moles of lactate = 4.123*10^-3 moles
Molarity of lactate = 4.123*10^-3 moles/0.05 L = 0.082 M
(Lactate)^- + H2O <==> HLac + OH-
Kb of lactic acid = Kw/ka = 1*10^-14/1.38*10^-4 =0.72 *10^-10
Kb = [HLac][OH-]/[Lac^-]
or, 0.72 *10^-10 = x^2/0.082-x
or, x = 2.4*10^-6
[Oh-] = 2.4*10^-6
pOH = -log (2.4*10^-6) = 5.62
pH = 8.38
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(c)
moles of NaOh added = 0.025 L 8 0.23 M = 5.75*10^-3 moles
extra moles of NaOH = (5.75*10^-3 moles) - (4.123*10^-3 moles) = 1.63 *10^-3 moles
Molarity = 1.63 *10^-3 moles /total volume = 0.033 M
pOH = 1.48
pH = 14-pOH = 12.52
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pH = pKa + log [lactate/lactic acid]
= 3.86 + log [5.75*10^-3 moles/4.123 *10^-3] = 4
HLa H+ La^- inital 0.0825 0 0 change -x _x +x equilibrium 0.0825-x +x +xRelated Questions
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