-If 35.00 mL of 0.0200 M aqueous HCl is required to titrate
30.00 mL of an aqueous solution of NaOH to the equivalence point,
the molarity of the NaOH solution is Blank 1 M.
-20.00 mL of 0.510 M NaOH is titrated with 0.740 M
H2SO4. Blank 1 mL of
H2SO4 are needed to reach the end point.
A 71.0 mL sample of 0.0400 M HIO4 is titrated with 0.0800 M RbOH
solution. Calculate the pH after the following volumes of base have
been added.
(a) 10.3 mL
pH =
(b) 34.4 mL
pH =
(c) 35.5 mL
pH =
(d) 36.2 mL
pH =
(e) 63.9 mL
pH =
A 83.0 mL sample of 0.0200 M HIO4 is titrated with 0.0400 M CsOH
solution. Calculate the pH after the following volumes of base have
been added. (
a) 11.6 mL
pH =
(b) 40.7 mL
pH =
(c) 41.5 mL
pH =
(d) 43.6 mL
pH =
(e) 75.5 mL
pH =
A 77.0 mL sample of 0.0400 M
HNO3 is titrated with 0.0800 M
CsOH solution. Calculate the pH after the following
volumes of base have been added.
(a) 10.0
mL
pH =
(b) 37.3
mL
pH =
(c) 38.5
mL
pH =
(d) 39.3
mL
pH =
(e) 65.1
mL
pH =
If 9.89 mL of 0.106 M sodium hydroxide is required to titrate
the acetylsalicylic acid in an aspirin tablet, how many milligrams
of acetylsalicylic acid are in the tablet? (MW acetylsalicylic acid
= 180.157 g/mol)
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 0.400 g sample of toothpaste was boiled with a 25 ml solution
TISAB buffer to extract the fluoride. After cooling, the solution
was diluted to exactly 100.0ml with more TISAB solution. The
potential of fluoride ion-selective electrode/reference electrode
in an aliquot of the sample was measured to be 186.6 mV. This value
corresponding to a log[F-] of -3.5016 on a working curve of mV
versus log [F-]. Calculate the weight percent F- and NaF in the
toothpaste sample.
100 mL of a river water sample took 9.30 mL of 0.01005 M Ag+ to
titrate. Calculate the concentration of Cl- in ppm (μg/mL or mg/L)
for the river water.
Part A: How many milliliters of 0.120 M NaOH are required to
titrate 50.0 mL of 0.0998 M butanoic acid to the equivalence point?
The Ka of butanoic acid is 1.5 x 10^-5.
Part B: What is the pH at the equivalence point?