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chemistry module 5 > acids,bases and buffers > Flashcards

acids,bases and buffers Flashcards

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1
Q

What is meant by a weak acid and Brønsted–Lowry acid

A

Proton/H+ donor
AND
Partially dissociates/ionises

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2
Q

Calculate the pH of 0.500 mol dm–3 potassium hydroxide.

A 
H+] = 1.00 ×10–14 0.5(00) OR 2(.00) × 10–14 (mol dm–3) 
pH = –log 2(.00) × 10–14 = 13.7(0) 
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3
Q

Write the full equation for the reaction of aqueous propanoic acid with sodium carbonate.

A

2C2H5COOH + Na2CO3 → 2C2H5COONa + CO2 + H2O

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4
Q

Write the ionic equation for the reaction of aqueous propanoic acid with aqueous potassium hydroxide.

A

H+ + OH– → H2O

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5
Q

A small amount of aqueous ammonia, NH3(aq), is added to the buffer solution.Explain, in terms of equilibrium, how the buffer solution would respond to the added NH3(aq).

A 
Added ammonia
C2H5COOH removes added NH3/alkali/base
OR C2H5COOH + NH3 / OH– →
OR NH3/alkali reacts with/accepts H+
OR H+ + NH3 →
OR H+ + OH– → 
Equlibrium → C2H5COO– OR Equilibrium → right 
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6
Q

How can an aqueous solution of an acid contain hydroxide ions?

A 
Water dissociates/ionises
OR
H2O ⇌ H
\+
\+ OH
–
OR
2H2O ⇌ H3O
\+
\+ OH– 
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7
Q

Write a full equation for the reaction between ethanoic acid and solid calcium carbonate.

A

2CH3COOH + CaCO3 → (CH3COO)2Ca + CO2 + H2O

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8
Q

Explain why this buffer solution has formed.

A

solution contains CH3COOH AND CH3COO– 

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9
Q

Explain how this buffer solution controls pH when either an acid or an alkali is added.

A

Quality of written communication, QWC
2 marks are available for explaining how the equilibrium
system allows the buffer solution to control the pH on addition
of H+
and OH-
(see below)
——————————————————
CH3COOH ⇌ H
+
+ CH3COO–

—————————————————–
CH3COOH reacts with added alkali
OR CH3COOH + OH– 
OR added alkali reacts with H+
OR H
+
+ OH–  
Equilibrium  right OR Equilibrium  CH3COO–  (QWC)
CH3COO–
reacts with added acid 
Equilibrium  left OR Equilibrium  CH3COOH  (QWC)

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10
Q

0.14 mol dm–3 solutions of hydrochloric acid, HCl, and chloric(I) acid, HClO (pKa = 7.43), have different pH values.Explain why the pH values are different and calculate the pH of 0.14 mol dm–3 solutions of HCl and HClO to two decimal places.
Show any working in calculations.

A 
HCl is a strong acid AND HClO is a weak acid

HCl:
pH = –log 0.14 = 0.85 (2 DP required)

HClO:
CHECK THE ANSWER ON ANSWER LINE
IF answer = 4.14, award all three calculation marks
--------------------------------------------------------------------
Ka = 10–7.43 OR 3.7 x 10–8 (mol dm–3) 
[H+] =
Ka  [HClO] OR
Ka  [HA]
OR
Ka  0.14 OR 3.7 x 108 x 0.14

pH = 4.14 (2 DP required)

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11
Q

Aluminium powder is added to aqueous ethanoic acid, CH3COOH.Write full and ionic equations for the reaction that takes place.

A

2Al + 6CH3COOH  2(CH3COO)3Al + 3H2 

2Al + 6H+  2Al3+ + 3H2 

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12
Q

Calculate the pH of a 0.40 mol dm–3 solution of NaOH.

A 
[H+] =
Kw
[OH ]
OR 1.0
 1014
[OH ]
OR 1.0
 1014
0.4(0)
OR 2.5 x 10–14 (mol dm–3) 
Correctly calculates pH = –log 2.5 x 10–14 = 13.6(0)

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13
Q

Explain what is meant by the term buffer solution.

Describe how a buffer solution based on methanoic acid can act as a buffer.

A 
A buffer solution minimises pH changes

on addition of small amounts of acid/H+ or alkali/OH–/base

--------------------------------------------------------------------------------
HCOOH
⇌ H+ + HCOO– 
Equilibrium sign essential 
Added alkali
HCOOH reacts with added alkali/base/OH–
OR added alkali/OH– reacts with H+ 
QWC: Equilibrium shifts forming HCOO– OR H+
OR (HCOOH) Equilibrium
 right

Added acid
HCOO– reacts with added acid/H+ 
QWC: Equilibrium shifts forming HCOOH
OR (HCOOH) Equilibrium
 left

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14
Q

A chemist prepares a buffer solution by mixing together the following:200 cm3 of 3.20 mol dm–3 HCOOH (Ka = 1.70 × 10–4 mol dm–3) and800 cm3 of 0.500 mol dm–3 NaOH.The volume of the buffer solution is 1.00 dm3. • Explain why a buffer solution is formed when these two solutions are mixed together. • Calculate the pH of this buffer solution. Give your answer to two decimal places.

A 
HCOOH reacts with NaOH forming HCOO–
/HCOONa
OR
HCOOH + NaOH
 HCOONa + H2O

Equilibrium sign allowed
(Some) HCOOH/(weak) acid remains
OR HCOOH/(weak) acid is in excess

n(HCOOH) OR [HCOOH]
= 0.24(0) (mol / mol dm–3) 
n(HCOO–) OR [HCOO–
] OR [HCOONa]
= 0.4(00) (mol / mol dm–3) 
 [H+] =
K a  [HCOOH] [HCOO– ]

pH = –log [H+] = –log(1.70
10–4

0.24
0.4 ) = 3.99

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15
Q

Write the ionic equation for the reaction between aqueous butanoic acid and magnesium.

A

Mg + 2H+  Mg2+ + H2

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16
Q

Write the ionic equation for the reaction between aqueous butanoic acid and aqueous sodium carbonate.

A

CO32– + 2H+  H2O + CO2

17
Q

The student adds 50.0 cm3 of 0.250 mol dm–3 butanoic acid to 50.0 cm3 of 0.0500 mol dm–3sodium hydroxide. A buffer solution forms.
Explain why a buffer solution forms.

A

CH3(CH2)2COONa OR CH3(CH2)2COO– forms
OR
CH3(CH2)2COOH + OH–  CH3(CH2)2COO– + H2O 
CH3(CH2)2COOH is in excess OR acid is in excess
OR some acid remains

18
Q

What is the difference between a strong acid and a weak acid?

A

A strong acid completely dissociates
AND
a weak acid partially dissociates

19
Q

Calculate the pH of 0.375 mol dm–3 nitrous acid, HNO2.Give your answer to two decimal places.

A

pH = –log 0.0129 = 1.89 
OR
pH = –log 0.0129 = 1.9

20
Q

Explain what is meant by the term Brønsted–Lowry base.

A

Proton acceptor

21
Q

Calculate the pH of 0.0400 mol dm–3 Ca(OH)2.Give your answer to two decimal places.

A 
[OH–] = 2 × 0.04(00) = 0.08(00) (mol dm–3) 
[H+] = 1.00
 10–14
0.08(00)
OR 1.25 × 10–13 (mol dm–3) 
pH = –log 1.25 × 10–13 = 12.90

---------------------------------------------
pOH variation (also worth 3 marks)
[OH–] = 2 × 0.04(00) = 0.08(00) (mol dm–3) 
pOH –log 0.08(00) = 1.10

pH = 14.00 – 1.10 = 12.90

22
Q

Aqueous calcium hydroxide is added to nitrous acid, HNO2.Write the overall equation and the ionic equation for the reaction that takes place.

overall: ……………………………………………………………………………………………………………………..
ionic: ……………………………………………………………………………………………………………………

A

Ca(OH)2 + 2HNO2  Ca(NO2)2 + 2H2O

H+ + OH–  H2O

23
Q

Carbonic acid, H2CO3, is a weak Brønsted–Lowry acid formed when carbon dioxide dissolves in water. Healthy blood is buffered to a pH of 7.40. The most important buffer solution in blood is a mixture of carbonic acid and hydrogencarbonate ions, HCO3–.
Explain how the carbonic acid–hydrogencarbonate mixture acts as a buffer in the control of blood pH.
In your answer you should explain how equilibrium allows the buffer solution to control the pH.

A 
Equilibrium
H2CO3 ⇌ H+ + HCO3
– 
Action of buffer
Added alkali
H2CO3 reacts with added alkali
OR H2CO3 + OH– 
OR added alkali reacts with H+
OR H+ + OH– 

Equilibrium
 right
OR equilibrium shifts forming H+ OR HCO3
– 
Added acid
HCO3
– reacts with added acid

Equilibrium
 left
OR equilibrium shifts forming H2CO3 
24
Q

Healthy blood at a pH of 7.40 has a hydrogencarbonate : carbonic acid ratio of 10.5 : 1. A patient is admitted to hospital. The patient’s blood pH is measured as 7.20.Calculate the hydrogencarbonate : carbonic acid ratio in the patient’s blood.

A 
n blood at pH 7.40,
[H+] = 10–pH = 10–7.40 = 3.98 × 10–8 (mol dm–3) 
Ka = [H+ ] [HCO3
– ]
[H2CO3 ]
 =
3.98
 10–8
 10.5
1
OR
Ka = 4.18 × 10–7 (mol dm–3) 
In blood at pH 7.20,
[H+] = 10–pH = 10–7.20 = 6.31 × 10–8 (mol dm–3) 
[HCO3
–
]
[H2CO3 ]

Ka
[H+
]
OR 4.18
 107
6.31
10–8

=
6.6
1
OR 6.6 : 1
 (up to calc. value, see below)
25
Q

What is meant by the term Brønsted–Lowry acid ?

A

proton donor 

26
Q

What is meant by the strength of an acid?

A

(the proportion of) dissociation

27
Q

What important factor does the student need to consider when deciding on the most suitable indicator to use for this titration?

A

Vertical section matches the (pH) range (of the
indicator)
OR colour change (of the indicator)
OR end point (of the indicator) 

28
Q

explain whether the dissociation of water is an exothermic or endothermic process.

A

Endothermic because Kw increases with

temperature 

29
Q

Many experimental measurements use published data, such as Kw, measured at 25 °C. Often these measurements have been taken at different temperatures, especially in experimental work carried out at body temperature.What is the consequence of this for published scientific work?

A

(Work is) inaccurate OR inva
because K varies with temperature
lid
w 

30
Q

The reverse reaction of the dissociation of water is called neutralisation.Plan an experiment that a student could carry out to measure the enthalpy change of neutralisation.

A 
Acid and alkali mixed 
Amounts of acid AND alkali stated 
Temperature taken at start AND finish 
energy, Q = mc∆T OR in words
AND meaning of m, c AND ∆T given 
Energy scaled up to form 1 mol of water
∆H = –energy change
31
Q

A student made a buffer solution by mixing an excess of propanoic acid to an aqueous solution of sodium hydroxide at 25 °C. This buffer solution contains an equilibrium system that minimises changes in pH when small amounts of acids and alkalis are added. • Explain why a buffer solution formed when an excess of propanoic acid was mixed with aqueous sodium hydroxide.• Explain how this buffer solution controls pH when an acid or an alkali is added.

A 
Propanoic acid reacts with sodium hydroxide
forming propanoate ions/sodium propanoate
OR
CH3CH2COOH + NaOH
 CH3CH2COONa + H2O

Some propanoic acid remains
OR
propanoic acid AND propanoate (ions)
/ sodium propanoate present

equilibrium: CH3CH2COOH
⇌ H+ + CH3CH2COO–

Added alkali
CH3CH2COOH reacts with added alkali
OR CH3CH2COOH + OH– 
OR added alkali reacts with H+
OR H+ + OH– 

 CH3CH2COO– OR Equilibrium
 right

Added acid
CH3CH2COO– reacts with added acid
OR [H+] increases

 CH3CH2COOH OR Equilibrium
 left

32
Q

Write an equation for the reaction of aqueous propanoic acid with magnesium.

A

2CH3CH2COOH + Mg

 (CH3CH2COO)2Mg + H2 

33
Q

Write an ionic equation for the reaction of aqueous propanoic acid with aqueous sodium carbonate.

A 
2H+ + CO32–  H2O + CO2
OR 2H+ + CO32–  H2CO3
OR H+ + CO32–  HCO3
–


chemistry module 5 (8 decks)

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