Question

Consider the titration of 25.00 mL of 0.1250 M AgNO3 with 0.05012 M KI. Ksp, AgI(s) = 8.3*10-17. For the same titration as shown in Question 2, please calculate the volume of KI that is required to reach a pAg of 2.00.

Answer #1

Consider the titration of 25.00 mL of 0.1250 M AgNO3 with
0.05012 M KI. ksp = AgI(s) = 8.3*10-17.
1) For the same titration as shown in Question 2,
please calculate the pAg value after 90.00 mL of KI is added.
2) For the same titration as shown in Question 2,
please calculate the volume of KI that is required to reach a pAg
of 2.00.

Consider the titration of 25.00 mL of 0.1250 M AgNO3 with
0.05012 M KI.
Ksp, AgI(s) = 8.3*10-17.
please calculate the volume of KI that is required to reach a
pAg of 2.00.
Answer is 47.82 mL. How they got that?

a) A 0.1500 M of AgNO3 solution was employed to
titrate a 25.00 mL of 0.1250 M of NaI and 0.2500 M NaCl. Given that
Ksp, AgI(s) = 8.3*10-17, and Ksp, AgCl(s) =
1.8*10-10, please calculate the concentration of
Ag+ ion after 6.00 mL of AgNO3 was added.
b) please calculate the pAg after 100.00 mL of AgNO3
was added

a) A 0.1500 M of AgNO3 solution was employed to titrate a 25.00
mL of 0.1250 M of NaI and 0.2500 M NaCl. Given that Ksp, AgI(s) =
8.3*10-17, and Ksp, AgCl(s) = 1.8*10-10, please calculate the
concentration of Ag+ ion after 6.00 mL of AgNO3 was added.
Answer: 1.2 x 10-15 M
b) please calculate the pAg after 100.00 mL of AgNO3 was
added
Answer: 1.35
I just want to see how they got that.

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing
0.1000 M sodium chloride (NaCl) and 0.05000 M potassium iodide
(KI), what is the pAg of the solution after 15.00 mL of AgNO3 is
added to the solution? Ksp, AgCl (s) = 1.82 x 10-10; Ksp, AgI(s) =
8.3*10-17.
b. Same titration as in (a), what is the pAg of the solution
after 25.00 mL of AgNO3 is added to the above solution?
c. Same titration as...

At 25 °C, you conduct a titration of 15.00 mL of a 0.0460 M
AgNO3 solution with a 0.0230 M NaI solution within the following
cell: Saturated Calomel Electrode || Titration Solution | Ag
(s)
For the cell as written, what is the voltage after the addition
of the following volume of NaI solution? The reduction potential
for the saturated calomel electrode is E = 0.241 V. The standard
reduction potential for the reaction
Ag+ + e- --> Ag(s)
is...

If 0.040 mol AgNO3 and 0.10 mol KI are added to 200.0
mL of water, what concentration of Ag+ remains in the
solution once equilibrium is established? The Ksp of AgI
is 8.5 X 10-17.

Find the concentration of I– in 0.050 M AgNO3 saturated with
AgI. Include activity coefficients in your solubility-product
expression. The Ksp of AgI is 8.3× 10–17. A table of activity
coefficients at various ionic strengths can be found
here.(https://sites.google.com/site/chempendix/activity-coefficients-for-aqueous-solutions-at-25-c)

A group of students conducted a titration of 25.00-mL saturated
Ca(OH)2 solution with 0.0480M HCl. The students found
that 10.10-mL of acid was required to reach the equivalence point.
Calculate the molar concentration of OH- and
Ca2+, the molar solubility, and Ksp of the
analyte. Show detailed calculation for each.
[OH-] in the analyte:
[Ca2+] in the analyte:
molar solubility of Ca(OH)2:
Ksp of Ca(OH)2:

Consider the titration of 25.00 mL of 0.300 M H3 PO4 with 0.300
M NaOH.
A. Write the net ionic equation for the reaction of phosphoric
acid with NaOH to form sodium phosphate and water:

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