ANALYTICAL CHEMISTRY FINAL- ACS EXAM QUESTIONS AND ANSWERS A+ GRADED. BUY Quality Materials!
In a titration experiment, the change of indicator color corresponds to the:
endpoint
This is the value of the equilibrium constant (at 25 C) of the reaction involved in the titration of strong acid wi...
ANALYTICAL CHEMISTRY FINAL- ACS EXAM QUESTIONS
AND ANSWERS A+ GRADED. BUY Quality Materials!
In a titration experiment, the change of indicator color corresponds to the:
endpoint
This is the value of the equilibrium constant (at 25 C) of the reaction involved in
the titration of strong acid with a strong base:
1*10^14
HNO3 solution was standarized by titrating with 0.4541 g of
tris(hydroxymethyl)aminomethane (FM=121.14). A 35.37 ml volume of the titrant
was needed for complete titration. The molarity of the nitric acid solution was
this:
0.4541/121.14 = 0.003749 (mass)
molarity = mass/volume
0.003749/0.03537= 0.106
A titration reaction is assumed to have this point when the quantity of titrant
added is the exact amount necessary for stoichometric reaction with the target
analyte:
equivalance point
Phenolphthalein is widely used as an indicator in acid-base titration. This will be
the color of this indicator at pH of 7:
orange-red
In an acid-base titration experiment, a 75.00 mL volume of 0.13 M ammonium
hydroxide solution was titrated with a 0.16 M HCl. This system will no longer
represent a buffer system after the added volume of the titration exceeds this
threshold value:
75(0.13)=x(0.16)
x=60.94
when a weak acid is titrated with a strong base, this system is created as soon as
the first drop of the titrant is added to the sample:
buffer
This base represents a primary standard in acid-base titration:
Na2CO3, (CH2OH)3-CNH2, KHC8H4O4, KH(IO3)2
When a weak acid is titrated with a strong base, this will be the pH of the system
at the completion of the reaction:
Baisc
A titration curve representing neutralization of HCl with NaOH will contain this
number of inflection points:
one
A 0.05 M solution of formic acid (Ka= 1.80*10^-4) was titrated with a 0.1 M solution
of NaOH at 25 C, the equilibrium constant for the titration reaction has this value:
, 14-4=10
1.80 * 10^10
A 0.1 M aqueous solution of HCl was titrated with a 0.1 M aqueous solution of
NaOH. In this experiment, the equivalance point will be reached at this pH value:
pH=7.0
In a titration experiment, 50 mL of 0.1 M acetic acid (Ka=1.75 *10^-5) was titrated
with a 0.1 M NaOH. The pH of the solution at 30 mL of this titration will be this:
mol acetic acid = 0.1 M * 50 mL = 5 mmol
mol(NaOH) = 0.1 M * 30 mL = 3 mmol
Volume of Solution = 50 + 30 = 80 mL
pka=-log(ka)
= -log(1.75*10^-5)
= 4.75
PH = pka + log (salt/acid)
= 4.75 + log(3/5)
= 4.53
An analytical experiment in which the titration procedures is carried out with the
matrix that does not contain the target analyte is called this:
Blank titration
In a titration experiment, 50 mL of 0.1 M acetic acid (Ka=1.75 *10^-3) was titrated
with a 0.1 M NaOH. The pH of the solution at 50 mL of this titration will be this
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 50 mL = 5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 50 mL = 5 mmol
Volume of Solution = 50 + 50 = 100 mL
Kb of CH3COO- = Kw/Ka = 110^-14/1.7510^-5 = 5.714*10^-10
concentration of CH3COO-,c = 5 mmol/100 mL = 0.05M
Kb = x*x/(c)
x = sqrt (Kb*c)
x = sqrt ((5.71410^-10)0.05) = 5.345*10^-6
[OH-] = x = 5.345*10^-6
use:
pOH = -log [OH-]
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