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Chem 123 Lab Final UBC Actual Exam
Questions And Answers Practice Questions
with SolutionS ALREADY GRADED A+,
Exams of Nursing
Calculate the reaction rate at absorbance 0.50 in Experiment 2. (make sure to look at the units of
the graph if already in M-1, follow these steps). - --ANS---1. do abs/ slope 0.5/34380, this gives
concentration of the dye
2. find the bleach initial concentration (0.19M *0.40mL)/10mL
3. use the earlier calculated k value.
What do you expect the reaction rate to be when the dye concentration is half of the
concentration in the question above - --ANS---The rate should be half the rate, because it is a
pseudo first order reaction
An unknown sample of Yellow 6 solution has the absorbance equal to 0.44. The slope of the
Beer's law calibration plot constructed using a 3.4E-5M stock solution of Yellow 6 is 0.0117(%)-
1. What is the molar concentration of the unknown sample (in mole L-1)? - --ANS---2 ways to
answer
1. keep in % do 0.44/(0.0117), this gives the % concentration, covert this to decimal form
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2. multiply the decimal form by the stock concentration
1. change slope to M^-1, by multiplying by 100% and dividing by stock concentration, do 0.44/
new slope units and you will get the unknown dye right away.
exp 12 calculations - --ANS---ka= [H30+][A-/HA], (you can also use henderson hasselbach)
Consider the following two solutions:
Solution A: Buffer with [H2PO4-] = 0.10 M [HPO4=] = 0.20 M
Solution B: Buffer with [H2PO4-] = 0.15 M [HPO4=] = 0.30 M
What is true about the pH of these solutions?
Both solutions have the same pH
Which solution will be able to neutralize addition of 0.11M OH-?
Solution B
Which solution will be able to neutralize addition of 0.11M H3O+?
Both solutions
What is true about the buffer capacities of these solutions?
Solution B has a higher capacity - --ANS---Buffer capacity relates to the amount of acid or base
a buffer solution can absorb. The buffer solution with the most buffering agents ([HA] and [A-])
can absorb the most added acid or base. (the one with a larger volume of HA and A will have the
higher buffer capacity.
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The "buffer capacity" quantifies the ability of a buffer solution to absorb strong acid or base. The
number of moles of strong acid per litre of buffer than can be absorbed while lowering the pH by
only one unit has been defined as the acid buffer capacity of a solution. This concept allows us to
quantitatively compare different buffer solutions. Similarly, the number of moles of strong base
per litre of buffer that can be absorbed while raising the pH by only one unit is known as the base
buffer capacity of a solution - --ANS---
The pH of a 25.0 mL sample of a buffer changed by one unit when 4.06 mL of 0.350 M NaOH
was added to it. What is the base buffer capacity of this buffer? - --ANS---(base volume x base
concentration)/(total volume)
ratio of HA/A = H30=/Ka - --ANS---
You work in a chemistry lab, and are asked to prepare 250 mL of a H2PO4- / HPO4= buffer in
which the concentration of the weak acid component is 0.0316 M and the concentration of the
conjugate base is 0.0565 M. Your starting materials are 0.527 M H2PO4-, 1.00 M NaOH, and
distilled water. What volume (in mL) of the acid (the 0.527 M H2PO4-) do you need to prepare
this buffer solution? - --ANS---begin by finding the moles using the desired concentrations and
250mL
add the acid and base moles to get the starting moles for the acid, the moles for the base stay the
same
divide the mols of acid by the stating material concentration of 0.527 and use the mols of base
and divide by 1M, times by 1000 to get in mL.
Chem 123 Lab Final UBC Actual Exam
Questions And Answers Practice Questions
with SolutionS ALREADY GRADED A+,
Exams of Nursing
Calculate the reaction rate at absorbance 0.50 in Experiment 2. (make sure to look at the units of
the graph if already in M-1, follow these steps). - --ANS---1. do abs/ slope 0.5/34380, this gives
concentration of the dye
2. find the bleach initial concentration (0.19M *0.40mL)/10mL
3. use the earlier calculated k value.
What do you expect the reaction rate to be when the dye concentration is half of the
concentration in the question above - --ANS---The rate should be half the rate, because it is a
pseudo first order reaction
An unknown sample of Yellow 6 solution has the absorbance equal to 0.44. The slope of the
Beer's law calibration plot constructed using a 3.4E-5M stock solution of Yellow 6 is 0.0117(%)-
1. What is the molar concentration of the unknown sample (in mole L-1)? - --ANS---2 ways to
answer
1. keep in % do 0.44/(0.0117), this gives the % concentration, covert this to decimal form
,2|Page
2. multiply the decimal form by the stock concentration
1. change slope to M^-1, by multiplying by 100% and dividing by stock concentration, do 0.44/
new slope units and you will get the unknown dye right away.
exp 12 calculations - --ANS---ka= [H30+][A-/HA], (you can also use henderson hasselbach)
Consider the following two solutions:
Solution A: Buffer with [H2PO4-] = 0.10 M [HPO4=] = 0.20 M
Solution B: Buffer with [H2PO4-] = 0.15 M [HPO4=] = 0.30 M
What is true about the pH of these solutions?
Both solutions have the same pH
Which solution will be able to neutralize addition of 0.11M OH-?
Solution B
Which solution will be able to neutralize addition of 0.11M H3O+?
Both solutions
What is true about the buffer capacities of these solutions?
Solution B has a higher capacity - --ANS---Buffer capacity relates to the amount of acid or base
a buffer solution can absorb. The buffer solution with the most buffering agents ([HA] and [A-])
can absorb the most added acid or base. (the one with a larger volume of HA and A will have the
higher buffer capacity.
, 3|Page
The "buffer capacity" quantifies the ability of a buffer solution to absorb strong acid or base. The
number of moles of strong acid per litre of buffer than can be absorbed while lowering the pH by
only one unit has been defined as the acid buffer capacity of a solution. This concept allows us to
quantitatively compare different buffer solutions. Similarly, the number of moles of strong base
per litre of buffer that can be absorbed while raising the pH by only one unit is known as the base
buffer capacity of a solution - --ANS---
The pH of a 25.0 mL sample of a buffer changed by one unit when 4.06 mL of 0.350 M NaOH
was added to it. What is the base buffer capacity of this buffer? - --ANS---(base volume x base
concentration)/(total volume)
ratio of HA/A = H30=/Ka - --ANS---
You work in a chemistry lab, and are asked to prepare 250 mL of a H2PO4- / HPO4= buffer in
which the concentration of the weak acid component is 0.0316 M and the concentration of the
conjugate base is 0.0565 M. Your starting materials are 0.527 M H2PO4-, 1.00 M NaOH, and
distilled water. What volume (in mL) of the acid (the 0.527 M H2PO4-) do you need to prepare
this buffer solution? - --ANS---begin by finding the moles using the desired concentrations and
250mL
add the acid and base moles to get the starting moles for the acid, the moles for the base stay the
same
divide the mols of acid by the stating material concentration of 0.527 and use the mols of base
and divide by 1M, times by 1000 to get in mL.
