A 25.00 mL solution of 0.100 M CH3NH2 is titrated with 0.1331 M
HCl. What is...
A 25.00 mL solution of 0.100 M CH3NH2 is titrated with 0.1331 M
HCl. What is the pH at the equivalence point of the titration?
Species (K values) CH3NH2 (Kb = 3.74x10-4) CH3NH3+ (Ka =
2.70x10-11)
25.00 mL 0.100 M CH3NH2 is titrated with 0.100 M HCl. Calculate
the pH of this the titration solution after the addition of a) 12.5
mL and b) 25.00 mL of the titrant has been added. Based on these pH
values, select an appropriate indicator for the titration from
table at the end of lecture notes for Chapter 16.
A 10.0 mL solution of 0.300 M NH3 is titrated with a
0.100 M HCl solution. Calculate the pH after the addition
of 20.0 mL of the HCl solution.
(potentially useful info: Ka of NH4+ = 5.6 x 10−10)
A 25 mL aliquot of an HCl solution is titrated with 0.100 M
NaOH. The equivalence point is reached after 21.27 mL of the base
were added. Calculate 1) the concentration of the acid in the
original solution, 2) the pH of the original HCl solution and the
original NaOH solution, 3) the pH after 10.00 mL of NaOH have been
added, 4) the pH at the equivalence point, and 5) the pH after
25.00 mL of NaOH have been...
If 85.00 mL of 0.100 M HCl is mixed with 25.00 mL of 0.200 M
H2SO4 and 50.00 mL of 0.400 M NaOH. What is the resulting pH of the
final solution?
Please show all work and how the formulas are derived if
modified.
a. What volume of 0.085 M HCl is required to titrate 25.00 mL of
a 0.100 M NH3 solution to the equivalence point?
b. What is the pH at the equivalence point?
A 40.0 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH.
Calculate the pH after the addition of each of the following
volumes of NaOH: (a) 0.0 mL, (b) 10.0 mL, (c) 20.0 mL, (d) 40.0 mL,
(e) 60.0 mL. A plot of the pH of the solution as a function of the
volume of added titrant is known as a pH titration curve. Using the
available data points, plot the pH titration curve for the above...
25.0 mL of a 0.100 M NH3 is titrated with a strong acid. 0.100 M
HCl. Calculate the pH of the NH3 solution at the following points
during the titration: (Kb= 1.8 x 10^-5) A. Prior to the addition of
any HCl. B: After the addition of 10.5 mL of a 0.100 M HCl. C: At
the equivilance point. D: After the addition of 3 mL of 0.100 M
HCl. Show your work.
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M
NaOH solution. Calculate the pH at each of the following
points.
Volume of NaOH added: 0, 5, 10, 25, 40 ,45 , 50 , 55 , 60 , 70 ,
80 , 90 , 100
25.0 mL of a 0.100 M HCN solution is titrated with
0.100 M KOH. Ka of HCN=6.2 x 10^-10.
a)What is the pH when 25mL of KOH is added (this is the equivalent
point).
b) what is the pH when 35 mL of KOH is added.
A 20.00 mL solution of 0.100 M HCOOH (formic) was titrated with
0.100 M KOH. The Ka for the weak acid formic is 1.40 x 10-5. a.
Determine the pH for the formic prior to its titration with KOH. b.
Determine the pH of this solution at the ½ neutralization point of
the titration. c. Identify the conjugate acid-base pair species at
the ½ neutralization point.