A) C2H3O(B); H20 (A); OH (CA); HC2Hs02 (CB) HCaHO2 K-5.6 x 10-10 B) C2H30f (CB);
ID: 1044021 • Letter: A
Question
A) C2H3O(B); H20 (A); OH (CA); HC2Hs02 (CB) HCaHO2 K-5.6 x 10-10 B) C2H30f (CB); H2O (CA); OH-(A); HCII,02"(B) C2H302(A); H20 (B): OH (CA); HC2H02 (CB) 2H3O24B); H20 (A); OH , (CB); HCthO2 (CA) 30. Determi he the pll at the equivalence point in the titration of 25.00 mL of 0.300 AM Lactic Acid (Ka 8.3 x 10- using 0.300 M NaOH A) 9.26 B)8.04 C)7.00 D) 8.13 31. Determine the pH at the hal titration point during the titration of 25.00 mL. of 0.300 M Lactic Acid (Ka -8.3 x 10-4) using 0.300 M NaOH H20 (o HCsHsO3 (aq) + OH-(aq) ? CsH5O3-(aq) + C)3.86 D) 3.96 A) 3.25 3.08 3.86Explanation / Answer
30.
25 mL of 0.300 M HC3H5O3
Moles of HC3H5O3 = 0.300 M x 0.025 L = 0.0075 moles
At equivalence point moles of NaOH = 0.0075 moles
[NaOH] = 0.300 M
So, Volume of NaOH = (0.0075 mol) / (0.300 M)
= 0.025 L
= 25 mL
At equivalence point 0.0075 moles of HC3H5O3 will react with 0.0075 moles of NaOH to form 0.0075 moles of NaC3H5O3.
Total volume = 25 mL + 25 mL = 50 mL = 0.050 L
[NaC3H5O3] = (0.0075 moles) / (0.050 L) = 0.15 M
C3H5O3- + H2O <===> HC3H5O3 + OH-
IC: 0.15 0 0
C: - x + x + x
EC: 0.15 – x x x
So, Kb = [HC3H5O3] [OH-] / [C3H5O3-]
Ka = 8.3 x 10-4
Kb = Kw / Ka
= (1.0 x 10-14) / (8.3 x 10-4)
= 1.20 x 10-11
So,
1.20 x 10-11 = (x) (x) / (0.15 – x)
1.20 x 10-11 = x2 / (0.15 – x)
1.20 x 10-11 = x2 / 0.15 (Ka is very small, so, x term in denominator is neglected)
x2 = (1.20 x 10-11) x 0.15
x2 = 1.8 x 10-12
x = 1.34 x 10-6
So, [OH-] = x = 1.34 x 10-6
pOH = -log [OH-] = - log (1.34 x 10-6) = 5.87
pH = 14 – pOH
= 14 – 5.87
= 8.13
Answer is option (D)
31.
At half equivalence point, the concentration of HC3H5O3 and NaC3H5O3 are same.
Using Henderson-Hesselbalach equation
pH = pKa + log { [NaC3H5O3] / [HC3H5O3]}
pH = pKa + log (1) [since, [NaC3H5O3] = [HC3H5O3]
pH = pKa
Ka = 8.3 x 10-4
pKa = - log Ka
= - log (8.3 x 10-4)
= 3.08
So, pH = pKa = 3.08
Answer is option (B)
33.
[NaC3H5O3] = (0.0075 moles) / (0.050 L) = 0.15 M
C3H5O3- + H2O <===> HC3H5O3 + OH-
IC: 0.15 0 0
C: - x + x + x
EC: 0.15 – x x x
So, Kb = [HC3H5O3] [OH-] / [C3H5O3-]
Ka = 8.3 x 10-4
Kb = Kw / Ka
= (1.0 x 10-14) / (8.3 x 10-4)
= 1.20 x 10-11
So,
1.20 x 10-11 = (x) (x) / (0.15 – x)
1.20 x 10-11 = x2 / (0.15 – x)
1.20 x 10-11 = x2 / 0.15 (Ka is very small, so, x term in denominator is neglected)
x2 = (1.20 x 10-11) x 0.15
x2 = 1.8 x 10-12
x = 1.34 x 10-6
So, [OH-] = x = 1.34 x 10-6
pOH = -log [OH-] = - log (1.34 x 10-6) = 5.87
pH = 14 – pOH
= 14 – 5.87
= 8.13
Answer is option (D)
34.
Answer is option (A)
An Arrhenius acid is a substance that dissociates in water to form hydrogen ions (H+).
35.
Answer is option (D)
An Arrhenius base is a substance that, when dissolved in an aqueous solution, increases the concentration of hydroxide, or OH-, ions in the solution.
36.
Answer is option (A)
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