Question

(a) If your titration solution is 0.410 M in NaOH, and the endpoint occurs at 13.50 mL of titrant, how many mmol of NaOH are required to reach the endpoint?

__?__mmol NaOH

(b) How many mmol of acetic acid
(HC_{2}H_{3}O_{2}) are required to react
with the NaOH?

__?__ mmol HC_{2}H_{3}O_{2}

(c) How many grams of acetic acid is this?

__?__ grams HC_{2}H_{3}O_{2}

(d) If the mass of analyte is 10.05 grams, what is the mass % of
acetic acid in the analyte?

__?__ %

Answer #1

a)

mmol of NaOH = M(NaOH) * Volume in mL

= 0.410 M * 13.50 mL

= 5.535 mmol

Answer: 5.54 mmol

b)

mol of HC2H3O2 = mol of NaoH

So,

mmol of HC2H3O2 = 5.535 mmol

Answer: 5.54 mmol

c)

Molar mass of HC2H3O2,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of HC2H3O2,

m = number of mol * molar mass

= 5.535*10^-3 mol * 60.05 g/mol

= 0.3324 g

Answer: 0.332 g

d)

mass % = mass of HC2H3O2 * 100 / total mass

= 0.332 * 100 / 10.05

= 3.30 %

Answer: 3.30 %

a) If your titration solution is 0.409 M in NaOH, and the
endpoint occues at 13.00 mL of titrant, how many mmol of NaOH are
required to reach the endpoint?
____ mmol NaOH
b) How many mmol of acetic acid (HC2H3O2) are required to react
with the NaOH?
____ mmol HC2H3O2
c) How many grams of acetic acid is this?
____ grams HC2H3O2
d) If the mass of analyte is 10.15 grams, what is the mass % of
acetic acid...

After standardizing your NaOH solution, you have determined that
the concentration is actually 0.1125 M. What is the weight percent
concentration of KHP (MW=204.23) in your unknown sample of mass
1.625 g if it takes 38.40 mL of titrant to reach the equivalence
point?
A student weighs out 1.7500 grams of dried potassium hydrogen
phthalate standard (MW=204.23) and finds that it takes 39.05 mL to
reach the endpoint of the titration with the solution of NaOH. What
is the molarity...

A NaOH solution is standardized using KHP and the molarity of
the NaOH solution is determined to be 0.4150 M. Since the acid in
vinegar is the monoprotic acid, acetic acid, the concentration of
the acetic acid can easily be determined by titration. If 90.72 mL
of this solution is required to titrate 10.32 mL of vinegar to the
Phenolphthalein endpoint, what is the concentration of acetic acid
in the vinegar?
_______________ M
95.71 mL of NaOH solution is required...

if 9.76 mL of an acetic Acid solution required 9.80 mL of 1.0962
M NaOH to reach the endpoint, whats the concentration of the Acetic
Avid Solution?

A
sample of acetic acid (weak acid) was neutralized with .05M NaOH
solution by titration. 34 mL of NaOH had been used. Show your work.
CH3COOH + NaOH-----CH3COONa + H2O
a) Calculate how many moles of NaOH were used?
b) How many moles of aspirin were in a sample?
c) Calculate how many grams of acetic acid were in the
sample
d) When acetic acid is titrated with NaOH solution what is the
pH at the equivalence point? Circle the...

if 10.57 mL of an acetic Acid solution required 10.54 mL 0.9607
M NaOH to reach the endpoint, whats the concentration?

A 20.00-mL sample of formic acid (HCO2H) is titrated with a
0.100 M solution of NaOH. To reach the endpoint of the titration,
30.00 mL of NaOH solution is required. Ka = 1.8 x
10-4
What is the concentration of formic acid in the original
solution?
What is the pH of the formic acid solution before the titration
begins (before the addition of any NaOH)?

How many moles of acetic acid will react with 18.25 ml of 0.7500
M NaOH?
Calculate the mass (in grams) of acetic acid in 10.00 ml of a
solution which contains 0.0083 moles of acetic acid.
Assuming that the 10.00 ml sample has an overall mass of 10.00g.
Based on the above answer, calculate the percent of this total mass
which is due to acetic acid.

A 20.00-mL sample of formic acid (HCO2H) is titrated with a
0.100 M solution of NaOH. To reach the endpoint of the titration,
30.00 mL of NaOH solution is required. Ka = 1.8 x
10-4
What is the pH of the solution after the addition of 10.00 mL of
NaOH solution?
What is the pH at the midpoint of the titration?
What is the pH at the equivalence point?

The
titration of 0.02500 L of a diprotic acid solution with 0.1000 M
NaOH requires 34.72 mL of titrant to reach the second equivalence
point. The pH is 3.95 at the first equivalence point and 9.27 at
the second equivalence point.
What is the pKa1 and pKa2 of the acid?
Thank You

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 8 minutes ago

asked 8 minutes ago

asked 12 minutes ago

asked 18 minutes ago

asked 49 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago