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A 28.6 mL solution of HBr (a strong acid) of unknown concentration was titrated using 0.3194 mol L-1
NaOH using phenolphthalein as a pH indicator. At the moment when the solution turned a consistent
light pink colour, the burette volume read 28.12 mL. The initial burette reading before the experiment
was 7.58 mL. What was the concentration of HBr in the original HBr solution in mol L-1 ?
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To find the concentration of HBr in the original solution, you can use the concept of titration. In a
titration, the moles of the titrant (NaOH, in this case) reacting with the analyte (HBr) can be determined
based on the balanced chemical equation. Then, you can use this information to calculate the
concentration of HBr.
Here's how you can calculate it step by step:
Calculate the moles of NaOH used:
Moles of NaOH = (final burette reading - initial burette reading) × NaOH concentration
Moles of NaOH = (28.12 mL - 7.58 mL) × 0.3194 mol/L
Moles of NaOH = 20.54 mL × 0.3194 mol/L (convert mL to L by dividing by 1000)
Moles of NaOH = 0.00656836 mol
Since HBr and NaOH react in a 1:1 ratio according to the balanced chemical equation:
HBr + NaOH → NaBr + H2O
The moles of HBr in the original solution are also 0.00656836 mol.
Calculate the concentration of HBr in the original solution:
Concentration (mol/L) = Moles of HBr / Volume (L)
Volume (L) = 28.6 mL / 1000 (convert to L) = 0.0286 L
Concentration (mol/L) = 0.00656836 mol / 0.0286 L ≈ 0.2294 mol/L
So, the concentration of HBr in the original solution is approximately 0.2294 mol/L.
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Related Questions
Please don't provide handwritten solution ....
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5:50 1
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Question 5 of 20
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Determine the mass of solid NaCH;COO that
must be dissolved in an existing 500.0 mL
solution of 0.200 M CH3COOH to form a buffer
with a pH equal to 5.00. The value of Ka for
CH-COОH is 1.8 х 10-5.
1
2
Let x represent the original concentration of
CH;COO in the water. Based on the given
values, set up the ICE table in order to
determine the unknown.
CH3COOH+ H20(1) =H;O*(aq) +CH3COO-(a
Initial (M)
Change (M)
Equilibrium
(M)
5 RESET
0.200
5.00
-5.00
1.0 x 10-9
-1.0 × 10-9
1.0 x 10-5
-1.0 x 10-5
1.8 x 10-5
-1.8 x 10-5
х+ 5.00
x - 5.00
x + 1.0 × 10-9
х - 1.0 х 10-9
1.0 x 10-5
x - 1.0 × 10-5
x + 1.8 × 10-5
х - 1.8 х 10-5
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CaCO3(s) + 2H+ ⟶⟶ Ca2+ + CO2↑↑ + H2O
The excess acid requied 39.96 mL of 0.1004 M NaOH for complete titration:
OH- + H+ ⟶⟶ H2O
Find the weight percent of calcite in the limestone.
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Can you answer this and explain how you did it please?
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You are studying a clear solution and you added the pH indicator methyl violet. The
colour range of the indicator methyl violet in a clear solution when changing from
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You initial pH of the solution when tested with a pH meter is O.2. You are going to
add 250 drops of 0.1 M HCI.
Please select the most appropriate answers to the following two questions.
What is the initial colour of the solution at pH 0.2?
What is the colour of the solution and what will the pH be after the addition of the
HCI?
Select 2 correct answer(s)
The colour of the solution after the addition of HCI will be clear and the pH will
be less than 0.2.
The colour of the solution after the addition of HCI will be violet and the pH will
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please be as detailed as possible
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A chemist titrates a 25.0 mL sample of 0.239 M benzoic acid (C6H5COOH) against a 0.100 M solution of NaOH.The overall reaction is shown by the equation below:
C6H5COOH(aq)+NaOH(aq)C6H5COONa(aq)+H2O(l)
The Ka value for benzoic acid is 6.28 × 10–5. Calculate the pH at the start of the titration, before any NaOH has been added. Give your answer correctly to two places after the decimal.
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Moles of NaOH?
Moles of HCl?
Moles of MgCO3?
Mass of MgCO3?
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Determine the differences in pH when a strong base is used to titrate a strong acid and a weak
acid. Consider the following reactions:
HC1
(ад)
+ NaOH
(aq)
→ NaCl
+ H2O a
(ад)
HC;H;O2 (aq) + NaOH
→ NaCH;O2 (ag) + H2O )
(аq)
Experiment 1:
0.2773 M NaOH is used to titrate 10.00 mL of 0.7198 M HC1
Experiment 2:
0.2773 M NaOH is used to titrate 10.00 mL of 0.7198 M HC:H;O2 (K. = 1.75 x 10*)
Because the volumes and concentrations are the same for both titrations, the volume of titrant
required to reach the equivalence point is also the same: 25.96 mL of NaOH.
Calculate the following:
1. The pH of both acids before any base has been added (initial pH).
2. The pH at the equivalence point for the HCH;O2 + NaOH titration.
3. The pH when 25.00 mL has been added in the HCl + NaOH titration
4. The pH when 8.00 mL has been added in the HC:H;O2 + NaOH titration,
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this sample with the enzyme acetylcholinesterase to convert all of the acetylcholine to choline and acetic acid. The acetic acid
dissociates to yield acetate and hydrogen ions. At the end of the incubation period, you measure the pH again and find that it has
decreased to 5.66. Assuming there was no buffer in the assay mixture, determine the number of nanomoles of acetylcholine in
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decimal) are significant when you raise a decimal number to a power.
CH,—C−O−CH,—CH,—*N—CH,
Acetylcholine
CH 3
acetycholine in original sample:
CH3
H₂O
CH3
HO–CH,—CH,—*N–CH, + CH,
Choline
CH3
CH, T
Acetate
O + H+
nmol
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C@HNH, (
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HCl to reach the equivalence point.
CH_NHz(aq) + HyO* (aq) + C,H,NH, (aq)+H,O(l)
a What was the concentration of aniline in the original solution?
Concentration 0115
✔ M
The equivalence point for a reaction is the point at which one reactant has been completely consumed by addition of another
reactant. Thus,
[C,HNH]
[H₂O+]
[OH-]-
=
0.155 mol HC1
1.00 L
1
29.2 mL
b What are the concentrations of
H₂O¹,
OH, and
C6H5NHs at the equivalence point?
M
C
= 0.115 M
x 21.61 ml x
M
Correct
1 mol C, H, NH,
1 mol HCl
Show Hint
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(a) The amount of titrant would be the same because the moles of analyte was not changed.(b) The amount of titrant would be doubled because the analyte solution volume was doubled.(c) The amount of titrant would be halved because the was diluted by half.
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esc
V
An analytical chemist weighs out 0.344 g of an unknown diprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. She then
titrates this solution with 0.0700M NaOH solution. When the titration reaches the equivalence point, the chemist finds she has added 65.5 mL of NaOH
solution.
Calculate the molar mass of the unknown acid. Round your answer to 3 significant digits.
g
0x10
mol
X
3
?
Explanation
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D
C Search or type URL
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Group of answer choices
No, the concentration of benzoic acid (C6H5COOH) in this buffer solution is not high enough.
Yes, the concentration of benzoate ions (C6H5COO-) in this buffer solution is great enough.
Yes, the concentration of benzoic acid (C6H5COOH) in this buffer solution is great enough.
No, the concentration of benzoate ions (C6H5COO-) in this buffer solution is not high enough.
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dissociates to yield acetate and hydrogen ions. At the end of the incubation period, you measure the pH again and find that it has
decreased to 5.52. Assuming there was no buffer in the assay mixture, determine the number of nanomoles of acetylcholine in
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Acetylcholine
CH₂
acetycholine in original sample:
CH,—C−O−CH,—CH,AN-CH, HO–CH,—CH,*N–CH, + CHy
H₂O
I
CH3
CH3
Choline
CH₂
CH,-GO + H
Acetate
nmol
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