Calculate the pH of the solution that results when 20.0 mL of 0.1610 M glycolic
ID: 1016341 • Letter: C
Question
Calculate the pH of the solution that results when 20.0 mL of 0.1610 M glycolic acid is (This problem requires values in your textbook's specific appendices, which you can access through the OWL_v2 MindTap Reader. You should not use the OWL_v2 References' Tables to answer this question as the values will not match.) diluted to 55.0 mL with distilled water. pH = mixed with 35.0 mL of 0.0920 M NaOH solution. pH = mixed with 35.0 mL of 0.150 M NaOH solution. pH = mixed with 35.0 mL of 0.150 M sodium glycolate solution. pH =Explanation / Answer
Since I do not have access to the appendix (you couldn’t upload the same), I had to use the value of pKa from the internet.
Glycolic acid is a weak monoprotic acid having formula HOCH2CO2H, represented here as HA for simplicity. HA has pKa 3.83 at 25°C.
(a) 20.0 mL of 0.1610 M HA is diluted to 55.0 mL. Since HA is a monoprotic acid, we use the dilution equation as
V1*S1 = V2*S2 where V1 = 25.0 mL; S1 = 0.1610 M and V2 = 55.0 mL; therefore,
(20.0 mL)*(0.1610 M) = (55.0 mL)*S2
===> S2 = (20.0*0.1610)/55.0 M = 0.0585 M
HA dissociates as
HA (aq) <=====> H+ (aq) + A- (aq)
initial 0.0585 0 0
change - x + x + x
equilibrium (0.0585 – x) x x
Ka = [H+][A-]/[HA] = (x)(x)/(0.0585 – x)
We make an assumption here; x << 0.0585 so that (0.0585 – x) 0.0585; Also, pKa = 3.83 ===> Ka = 10-3.83 = 1.479*10-4
Thus, 1.479*10-4 = x2/0.0585
===> x2 = 8.652*10-6
===> x = 2.941*10-3
pH = -log10[H+] = -log10(2.941*10-3) = 2.53 (ans)
(b) 35.0 mL 0.0920 M NaOH solution is added. Moles of NaOH added = (35.0 mL)*(0.0920 mole/L) = 3.22 mmoles
Moles of acid present in the original solution = (20.0 mL)*(0.1610 mole/L) = 3.22 mmoles
This is the equivalence point. The reaction occurring here is
HA (aq) + NaOH (aq) <=====> H2O (l) + NaA (aq)
initial 3.22 3.22 0
change - 3.22 - 3.22 + 3.22
equilibrium 0 0 3.22
Molarity of NaA = (3.22 mmoles)/(20.0 mL + 35.0 mL) = 0.0585
NaA is the conjugate base of the weak acid, HA and will establish equilibrium as
A- (aq) + H2O (l) <=====> HA (aq) + OH- (aq)
initial 0.0585 0 0
change - x + x + x
equilibrium (0.0585 – x) x x
Also, since OH- is generated, we must use Kb, the base dissociation constant. Since, Ka = 1.479*10-4, therefore,
Kb = 1014/(1.479*10-4) = 6.761*10-11 = (x)(x)/(0.0585 – x)
Again, we employ the small x approximation as above and get
6.761*10-11 = x2/0.0585 ===> x2 = 3.955*10-12
===> x = 1.989*10-6
pOH = -log10[OH-] = -log10(1.989*10-6) = 5.70
pH = 14 – pOH = 14 – 5.70 = 8.30 (ans)
(c) 35.0 mL of 0.150 M NaOH is added. Moles of NaOH added = (35.0 mL)*(0.150 mole/L) = 5.25 mmoles
We write the neutralization reaction as
HA (aq) + NaOH (aq) <=====> H2O (l) + NaA (aq)
initial 3.22 5.25 0
change - 3.22 - 3.22 + 3.22
equilibrium 0 2.03 3.22
We now have two bases present, OH- and A-. OH- is a much stronger base than A- and hence the pH of the mixture will be governed by OH-. Therefore, molarity of OH- in the mixture = (2.03 mmoles)/(20.0 mL + 35.0 mL) = 0.0369 M
pOH = -log10[OH-] = -log10(0.0369) = 1.433
pOH = 14 – pOH = 14 – 1.433 = 12.567 12.57 (ans)
(d) We now have a buffer solution. Moles of sodium glycolate added = (35.0 mL)*(0.150 mol/L) = 5.25 mmoles
Total volume of the solution = (20.0 mL + 35.0 mL) = 55.0 mL
Molarity of HA = (3.22 mmoles)/(55.0 mL) = 0.0585 M (we calculated the initial molarity of HA in part (b) above).
Molarity of sodium glycolate = (5.25 mmoles)/(55.0 mL) = 0.0954 M
We use the Henderson-Hasselbach equation as
pH = pKa + log10[sodium glycolate]/[HA] = 3.83 + log10(0.0954 M/0.0585 M)
===> pH = pKa + log10(1.631) = 3.83 + 0.212 = 4.042 4.04 (ans)
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