Homework Answers
Hi there,
In the given figure they clearly mentioned at half-way to the equivalence point pH = pKa (dotted line and mentioning This pH = pKa) , this means at half-way point the concentration of weak acid = conjugate base, so that log[conjugate base / weak acid] term becomes zero in Henderson-Hasselbalch equation
pH = pKa + log[conjugate base / weak acid] = pka + 0 = pKa
pH = pKa
but Ka = 10^-pKa = 10^-pH = 10^-3.86 = 0.00013803842 = 1.38 x 10^-4
so Ka = 1.38 x 10^-4 (OR) simply in scientific notation 1.38E-04
Hope this helped you!
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can i get help with this question please
3. Consider the weak base-strong acid titration of 25.00 mL of 0.100 M NH, with 0.100 M HCl (weak base, strong acid). The Ks for this weak base is 1.8 x 10 a. Calculate the pH of the solution after the addition of 10.0 mL HCL. b. What is the pH half-way to the equivalence point? c. Calculate the pH at the equivalence point. d. Calculate the pH after the addition of...
Based on the following information...
please answer A. B. and C.
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The half‑equivalence point of a titration occurs half way to the
equivalence point, where half of the analyze has reacted to form
its conjugate, and the other half still remains unreacted.
If 0.4400.440 moles of a monoprotic weak acid
(?a=7.2×10−5)(Ka=7.2×10−5) is titrated with NaOH,NaOH, what is the
pH of the solution at the half‑equivalence point?
pH=pH=
2)
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to 565 mL565 mL of 0.250 M0.250 M weak acid...
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Answers can only be entered/graded between 2019-10-24 and 2019-10-29 A student peforms a titration, titrating 25.00 mL of a weak monoprotic acid, HA, with a 1.12 M solution of NaOH. They collect data, plot a titration curve and determine the values given in the below table. ml NaOH PH added Half-way Point 17.97 3.66 Equivalence point 35.938.65 How many moles of NaOH have been added at the equivalence point? r mol correct...
Using the following pH curve for the titration of a weak acid with a strong base, if the pH at half-equivalence point is 4.75, what is the Ka of the weak acid? Equivalence Point Half-equivalence Point - 8 12 14 Volume of base added (in ml) 20 1.78 x 10-4 1.77 x 10-4 1.77 x 10-5 1.78 x 10-5
Equivalence Point for Titration #1: 24.96
mL
Equivalence Point for Titration #2: 25.40
mL
Equivalence Point for Titration #3: 25.20
mL
Midpoint pH for Titration #3: 9.80
QUESTIONS:
4) Set up the calculation required to determine
the concentration of the NaOH solution via titration of a given
amount of KHP. Include all numbers except the given mass of
KHP.
5) Set up the calculation required to determine
the concentration of the unknown strong acid via titration with a
known volume...
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b. If the pH at the half equivalence point for a titration of a weak acid with a strong base is 4.6, what is the value of K for the weak acid?
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