A ) This titration curve best represents which of the following? The titration o
ID: 896714 • Letter: A
Question
A)
This titration curve best represents which of the following?
The titration of a strong acid with a strong base.
The titration of a strong base with a strong acid.
The titration of a weak acid with a strong base.
The titration of a weak base with a strong acid.
1 points
B)
25 mL of 0.20 M unknown acid is titrated with 0.20 M NaOH, giving the following titration curve. Which of these statements about this titration is true?
When 25 mL of strong base has been added, the resulting solution is a buffer.
The pKa of the weak acid is approximately 5.
The acid being titrated is a strong acid.
After 10 mL of strong base had been added, the pH is determined solely by the amount of weak base formed.
2 points
C)
35.0 mL of a weak acid was titrated to the equivalence point with 28.0 mL of 0.100 M NaOH. The pH was monitored during the titration. At 14.0 mL of base added, the pH was found to be 2.73. What is the Ka of the weak acid?
(NOTE: To enter answer in scientific notation, use E instead of x10. For example, 1.0x10-3 should be entered as 1.0E-3.)
1 points
D)
The weak base hydroxylamine, HONH2, is titrated with HCl. What is the pH after 9.05 mL of 0.20 M HCl is added to 30.0 mL of 0.20 M HONH2? Kb of HONH2 = 1.1 x 10-8 (Give your answer to two decimal places.)
a.The titration of a strong acid with a strong base.
b.The titration of a strong base with a strong acid.
c.The titration of a weak acid with a strong base.
d.The titration of a weak base with a strong acid.
Explanation / Answer
A) and B) image not visible
C) If we look at the Hendersen-Hasselbalck equation,
pH = pKa + log([base]/[acid])
If 28 ml of 0.1 M NaOH was needed to completely neutralize the weak acid [equivalence point]. 14 ml of 0.1 M NaOH addition is half-equivalence point
At half equivalence point, concentration of acid remaining = concentration of salt formed
Feeding in the Hendersen-Hasselbalck equation would give,
pH = pKa = 2.73 = -log[Ka]
Thus, Ka of weak acid = 1.86E-03
D) NH2OH titration with HCl
1 mole of base reacts with 1 mole of acid added
moles of HCl = molarity x volume = 0.2 M x 0.00905 L = 0.00181 mol
So, moles of NH2OH = 0.2 x 0.03 = 0.006 mol
excess moles of base remaining = 0.006-0.00181 = 0.0042 mols
molarity of base in solution = moles/total volume
= 0.0042/(0.00905 + 0.030) = 0.108 M
base reacts with H2O as,
NH2OH + H2O <===> HNH2OH+ + OH-
Kb = 1.1 x 10^-8 = [HNH2OH+][OH-]/[NH2OH]
let x amount of base reacted then,
1.1 x 10^-8 = x^2/0.108
x = [OH-] = 3.45 x 10^-5 M
pOH = -log[OH-] = 4.46
pH = 14 - pOH = 9.54
Thus, pH of solution is 9.54
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