CONNECTING THIS LAB TO THE LECTURE. Results must be reported to the correct numb
ID: 950801 • Letter: C
Question
CONNECTING THIS LAB TO THE LECTURE. Results must be reported to the correct number of significant figures. See Section 16.9. Ascorbic acid, H_2C_6H_6O_6, is a dicrotic acid usually known as vitamin C. For this acid, is 4.10 and pK, is 11.80. When 125 mL of a solution of ascorbic acid was evaporated to dryness, the residue of pure ascorbic acid had a mass of 3.12 g. Calculate the molar concentration of ascorbic acid in the solution before it was evaporated. Calculate the pH of the solution and the molar concentration of the acerbate ion, C_6H_6O_6^2, before the solution was evaporated. When 50.00 mL of an acid with a concentration of 0.115 M (for which pK_a = 4.87) is titrated with 0.100 M NaOH, what is the pH at the equivalence point? What would be a good indicator for this titration? Look for Table 16.7 to select the indicator. When 25.0 mL of 0.100 M NaOH was added to 50.0 mL of a 0.100 M solution of a weak acid, HX, the pH of the mixture reached a value of 3.56. What is the value of K_a for the weak acid?Explanation / Answer
1) Data
Molarity
M = mol/L
M = 3.12g/176.12g/mol * 1000/125 ml = 0.144
[H2C6H6O6] = 0.144 M
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pKa = 4.1
-log(pKa) = 4.1
Ka = 7.94E-5
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a) Concentration
Ka = [H+][HA-]/[H2A]
(7.94E-5) = x*x/(0.144)
x2 = 1.14E-5
x = [H+] = 3.88E-3 M
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b) pH.
pH = -log[h+]
pH = -log(3.88E-3)
pH = 2.47
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c) Titration
HA + NaOH <==> Na+ + A- + H2O
M1V1 = M2V2
0.115mol/L x 0.05 L = 0.100 mol/L x V2
V2 = 0.058 L = 58 ml NaOH
[A-] = 0.050 L x 0.115 mol/L / (0.050 + 0.058) L = 0.0535 M
A- + H2O <==> H+A- + [OH-]
kb = [HA] [OH-]/[A-] = kw/ka
[HA] = [OH-]
[OH-]2 = kb x [A-]
[OH-]2 = (1E-14/1E-4.94) x 0.0535
[OH-]2 = 4.65E-11
[OH-] = 6.82E-6 M
pOH = -log (6.82E-6)
pOH = 5.17
pH + pOH = 14
pH = 14 - 5.17
pH = 8.83
Indicator: Phenolphthalein (8.2 - 10.0)
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2) Data
Vt = 50.0 ml + 25.0 ml = 75.0 ml
moles NaOH = M x L = 0.100 mol/L x 0.025 L = 0.0025 mol
moles weak acid = 0.100 mol/L x 0.050 L = 0.0050 mol
H+A- + OH- <==> [A-] + H2O
0.005, 0.0025 -------0
(0.005-0.0025), 0 --0.0025
[HA] = [A-]
[A-] = 0.0025 mol/ 0.075 L = 0.0333 M
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Previous calculations are to determine the reaction, this solution will react as a buffer. So we can apply the following equation:
pH = pKa + log {[A-]/[HA]}
log {[A-]/[HA]} = 1
pH = pKa = 3.56
Ka = 10-3,56
Ka = 2.75 E-4
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