C.
B.
D.
C.
D.
2
NaOH
NaCl
Na 2 CO3
NaOH
Na 2 CO3
Acid Base Practice Test. 21. A basic solution
1 21. A
A.basic
tastes
sour.
solution
B. feels slippery.
A. does
tastesnot
sour.
C.
conduct electricity.
B.
feels
slippery.
D. reacts with metals to release oxygen gas.
C. does not conduct electricity.
D. reacts with metals to release oxygen gas.
2 22. The balanced formula equation for the neutralization of H 2SO 4 by KOH is
22. The
the neutralization of H 2SO 4 by KOH is
A. balanced
H 2SO 4 +formula
KOH equation
→ KSOfor
4 + H2O
B. H 2SO 4 + KOH → K 2SO 4 + H 2 O
A. Arrhenius
H 2SO 4 + base
KOHis defined
→ KSO
H2O
23. An
as 4a +substance
which
C. H 2SO 4 + 2KOH → K 2SO 4 + H 2 O
B. H 2SO 4 + KOH → K 2SO 4 + H 2 O
23. An
base
is defined
A. Arrhenius
donates
2KOH
→ K as
SOa substance
+ 2H 2 O which
D.
H SO +protons.
C. H22SO44 + 2KOH → K 22SO 44 + H 2 O
B. donates electrons.
3 23. An
A. Arrhenius
donates
protons.
2KOH
→ K asSO
+ 2H 2 O which
D.
H 2SO 4 +
base
is defined
a substance
C. produces
H + in solution.2 4
B. donates electrons.
A.
protons.
D. donates
produces
OH − in solution.
C. produces H + in solution.
B. donates electrons.
−
in solution.
D. produces OH
+
C. produces H in solution.
produces
OH − in solution.
24. D.
Consider
the following
equilibrium:
4 24. Consider the following equilibrium:
−
−
HS + H 3PO 4 →
← H 2S + H 2 PO 4
→ H 2S + H 2 PO 4 −
24. Consider the following equilibrium:
HS− + acids
H 3POand
4 ←
The order of Brönsted-Lowry
bases- 6is→
HS− + acids
H 3POand
H 2Sis+ H 2 PO 4 −
A. order
acid, base,
acid, base.
The
of Brönsted-Lowry
bases
4 ←
- 6B. acid, base, base, acid.
A.
acid,
base,
acid,
The
Brönsted-Lowry
C. order
base, of
acid,
acid, base.
base. acids and bases is
B.
acid,
base,
base,
D. base, acid, base, acid.
acid.
A.
C. acid,
base,base,
acid, acid,
acid, base.
base.
B.
D. acid,
base,base,
acid, base,
base, acid.
acid.
C. base, acid, acid, base.
5 25. D.
acid,representing
base, acid. the reaction of ethanoic acid with water is
The base,
equation
−
−
25. The
reaction
of ethanoic
acid with water is
+ H2O →
A. equation
CH 3COOrepresenting
←theCH
3COOH + OH
−
2−
+
→
− + H2O →
B. equation
CH 3COO
COO
CH
+ H 3O
− acid with water is
25. The
representing
theCH
reaction
←
2 COOof ethanoic
+ H2O ←
A.
CH
3
3COOH + OH
−
+
→
− +H O →
C. CH
CH 3COO
COOH
CH 3COO
2−+ H 3O +
← CH
− + H 22O →
B.
COO
+
H
O
−
←
3
2
3
A. CH 3COO + H 2 O ← CH 3COOH ++OH −
→
D. CH
CH 3COOH
COOH
+H O →
CH 3COO
COOH
− 2 + OH
+
← CH
− + H 22O →
C.
2− +
←
3
3
B. CH 3COO + H 2 O ← CH
+HH33OO +
2 COO
+
−
→
D. CH
CH 3COOH
COOH ++ H
H 2O
O→
CH 3COO
COOH
− 2 + OH
← CH
C.
+ H 3O +
←
3
2
3
+
−
→
CH 3COOH
+ H 2 Oequilibrium:
← CH 3COOH 2 + OH
26. D.
Consider
the following
+
−
26. Consider the following equilibrium:
2H 2 O( l ) + 57 kJ →
← H 3O ( aq ) + OH ( aq )
+
−
→
26. Consider the following equilibrium:
kJwater
← H 3O ( aq ) + OH ( aq )
2 O( l ) + 57
When the temperature is2H
decreased,
the
+
−
→
kJ water
2H 2 O
← H 3O ( aq ) + OH ( aq )
) + 57the
When
the temperature
O + ( lincreases.
A. stays
neutral and isHdecreased,
C.
D.
−
+
CH 3COOH + H 2 O →
← CH 3COO + H 3O
+
−
CH 3COOH + H 2 O →
← CH 3COOH 2 + OH
6 26. Consider the following equilibrium:
+
−
2H 2 O( l ) + 57 kJ →
← H 3O ( aq ) + OH ( aq )
When the temperature is decreased, the water
[ ]
B. stays neutral and [ H O ] decreases.
C. becomes basic and H O decreases.
27. In a solution at 25°C, the[ [H O] ] is 3.5 × 10 M. The [OH ] is
D. becomes acidic and [ H O ] increases.
27. A.
In a solution
the [H O ] is 3.5 × 10 M. The [OH ] is
M
3. 5 × 10 at 25°C,
7 27. B.
9 × 10 atM25°C, the H O is 3.5 × 10 M. The OH is
In a 2.
solution
[ ]
[ ]
A.
3.
5 × 10 M
A. stays neutral and H 3O + increases.
+
3
3
−20
−9
−20
−7
1. 0 × 10 −9M
2. 9 × 10−6−20M
3.3.55××10
10 MM
1. 0 × 10 −7−9M
2. 9 × 10 M
3. 5 × 10 −6−7M
C.
B.
D.
A.
C.
B.
D.
C. 1. 0 × 10 M
+
3
+
−6
−
+
−6
−
+
3
3
3
+
−6
OVER
- 7-
[ ]
8 28. In a solution with a pOH of 4.22, the OH is
[ ]
A. 1. 7 × 10 M
0 × 10 M a pOH of 4.22, the OH is
28. B.
In a 6.
solution
[ ]
M
A.
1.
7 × 10 with
−6
3. 5 × 10with
Ma pOH of 4.22, the OH − is
28. InD.a solution
−10
−5
−10
6.3 × 10 −1−5M
6. 0 × 104 −10M
1.1.77××10
10 M M
6.3 × 10 −1−5M
6. 0 × 10 M
1. 7 × 10 4−1
M
C.
B.
D.
A.
C.
B.
D.
C. 6.3 × 10 M
9 29. An aqueous solution
D. 1. 7 × 10 4 M of NH 4 CN is
−
−
−
29. A.
An aqueous
solutionKofa <NH
basic because
K b4 CN is
B. basic because K a > K b
basic because
K b CN is
29. A.
An aqueous
solutionKof
a <NH
C. acidic because K a < K b 4
B. basic because K a > K b
D.
A. acidic
basic because KK ><KK
C. acidic because Kaaa < Kbbb
B. basic because K a > K b
D. acidic because K a > K b
C. acidic because K a < K b
30. The
ionicbecause
equationKfor
D. net
acidic
K b predominant hydrolysis reaction of KHSO 4 is
a >the
− equation→
30. A.
The net
ionic
for SO
the42−
predominant
+ H 3O + hydrolysis reaction of KHSO 4 is
HSO
←
4 + H2O
−
→
H
Hthe
OH −+ hydrolysis reaction of KHSO is
2−
←
4− +
2O
2 SO
4 ++H
→
30. B.
The HSO
net ionic
equation
for
predominant
+
H
O
SO
O
A.
HSO
4
← +4
4
2
2−3
+
C. KHSO−4 + H 2 O →
K
+
SO
+
H
O
→
B. HSO 4 −+ H 2 O ←
H 2SO2−4 +4 OH −+ 3
←
→
+ 4
−
+
H
O
+H
O
A.
HSO
→
← KSO
D. KHSO44 + H 22O ←
2−43 + OH +
+ + H 2 SO
C. KHSO−4 + H 2 O →
K
+
SO
+
H
O
←
4
3
−
B. HSO 4 + H 2 O →
← H+2SO 4 + OH
→
D. KHSO 4 + H 2 O ←
K ++ H 2SO2−4 + OH −+
→
C. KHSO 4 + H 2 O ← K + SO 4 + H 3O
+
−
D. KHSO 4 + H 2 O →
← K + H 2SO 4 + OH
B. basic because K a > K b
C. acidic because K a < K b
D. acidic because K a > K b
10 30. The net ionic equation for the predominant hydrolysis reaction of KHSO 4 is
−
+
H2O →
SO 42− + Hfor
A. HSOthe
←equilibrium
4 +
3Oan indicator:
31. Consider
following
−
B. HSO 4− + H 2 O →
← H 2SO 4 + OH
+− + H O +
HInd+ + H 2 O2−→
3
C. KHSO 4 + H 2 O →
+ HInd
← K + SO
4 ←
3O
31. Consider the following →
equilibrium
for an indicator:
−
D. KHSO
+ H 2 O ←equilibrium
K + + H 2SO
+ OH
31. Consider
the4 following
for4 an
indicator:
At the transition
point,
−
→
HInd
+ H 2 O for
Indindicator:
+ H 3O +
11 ← an
31. Consider the following
equilibrium
−
→
HInd + H 2 O ← Ind − + H 3O +
A. [ HInd ] > Ind
At the transition point,
−
+
−
HInd + H 2 O →
B. the[ HInd
] = Ind
← Ind + H 3O
At
transition
point,
−−
A.
]]>< Ind
C. the[[HInd
HInd
Ind
At
transition
point,
A. [ HInd ] > Ind− −
+
B.
D. [[HInd
HInd]]== Ind
H 3O
−−
A.
B. [ HInd
HInd] >= Ind
Ind
C. [ HInd ] < Ind −
B.
C. HInd
HInd =< Ind −−
D. [ HInd ] = H O +
[ ]
[ ]
[[ ] ]
[[[ ] ] ]
[[ ]]
]]
[
] [[
The
C.
D. equivalence
HInd] <= [ Ind
Hpoint
O] ] in a titration is reached when 20.0 mL of H SO
[ HInd
to 20.0 mL of 0.420 M KOH. The [ H SO ] in the original solution is
D. [ HInd ] = [ H O ]
- 83
32.
32.
12 32.
32.
33.
13 33.
33.
33.
14 34.
34.
34.
34.
35.
35.
35.
3
3
−+
+
2
2
4
is added
4
The
H 2SO 4 is added
A. equivalence
0.00840 M point in a titration is reached when 20.0 mL of
original
is SO is added
to
mL M
of 0.420
M KOH.
The [ His2SO
4 ] in the
B.20.0
0.210
The
equivalence
point
in a titration
reached
when
20.0solution
mL of H
2
4
C. 0.420
to
mLM
ofM0.420 M KOH. The [ H 2SO 4 ] in the original solution is
A. 20.0
0.00840
The
equivalence
D. 0.840 M point in a titration is reached when 20.0 mL of H 2SO 4 is added
B.20.0
0.210
to
mLMofM
0.420 M KOH. The [ H 2SO 4 ] in the original solution is
A.
0.00840
C. 0.420 M
B. 0.840
0.210M
M
D.
A.
C. 0.00840
0.420 MM
B.
0.210
In
D. a titration
0.840 M
Mbetween a weak acid and a strong base, the pH at the equivalence point is
C. 0.420 M
A.a titration
3
D.
0.840
Mbetween a weak acid and a strong base, the pH at the equivalence point is
In
B. 5
C. a titration
A.
37
In
between a weak acid and a strong base, the pH at the equivalence point is
D. 59
B.
In
between a weak acid and a strong base, the pH at the equivalence point is
C.
73
A.a titration
D.
B. 95
A.
C. 37
B.
5
The
D. pH
9 of 100.0 mL of 0.0050 M NaOH solution is
C. 7
A. pH
The
of 100.0 mL of 0.0050 M NaOH solution is
D.
9 2.30
B. 3.30
A.
2.30
C. pH
10.70
The
of 100.0 mL of 0.0050 M NaOH solution is
B.
3.30
D. 11.70
C. pH
10.70
The
of 100.0 mL of 0.0050 M NaOH solution is
A. 2.30
D. 11.70
B. 3.30
A. 2.30
C. 10.70
B.
3.30solution is prepared by adding 1.0 mol of NaCH 3COO to 1.0 L of 1.0 M CH 3COOH.
A
D.buffer
11.70
−
C.
10.70
The
molar
concentration
of by
CHadding
approximately
A
buffer
solution
is prepared
of NaCH 3COO to 1.0 L of 1.0 M CH 3COOH.
3COO1.0ismol
D. 11.70
−
The molar concentration of CH 3COO is approximately
A. 0.0
A
B.buffer
0.5 solution is prepared by adding 1.0 mol of NaCH 3COO to 1.0 L of 1.0 M CH 3COOH.
A.
0.0
−
C. 1.0
A. 2.30
B. 3.30
C. 10.70
D. 11.70
15 [Cl– ]
[Cl– ]
C.
D.
35. A buffer solution is prepared by adding 1.0 mol of NaCH 3COO to 1.0 L of 1.0 M CH 3COOH.
+]
The molar
of CH 3COO − is approximately
[Ag+ ]
[Agconcentration
A.
B.
C.
D.
0.0
0.5
1.0
2.0
OVER
16 36. The equation for the reaction of Cl O -with
9 - water is
2
A.
B.
C.
D.
Cl 2 O + H 2 O
Cl 2 O + H 2 O
Cl 2 O + H 2 O
Cl 2 O + H 2 O
→
←
→
←
→
←
→
←
2HClO
–
2ClO + H[Cl
2 ]
Cl 2 + H 2 O 2
Cl 2 + O 2 + H 2
17 20. The acid found in vinegar will
37.
21.
21.
18 21.
22.
19 38.
22.
22.
20 23.
[Cl– ]
Consider
following redox reaction:
A. taste the
bitter.
In
which
of
the
following equilibrium systems is HCO3− acting as a Brönsted-Lowry base?
B. feel slippery.
−
+
2MnO
+ 2Mn 2+ + 3H 2 O
C. change− litmus
to
blue.
4 + 5CH
3CHO + 6H → 5CH 3COOH
−
2−
+
→
acting
as a Brönsted-Lowry base?
In
which
of
the
following
equilibrium
systems
is
HCO
A.
CO3 H
3
D. HCO
react 3with←MgHto +produce
2
− →electrons is 2−
The species
B.
HCO − that
+ HSloses
H S + CO3 A. HCO33− →
H +←+ CO232−
←
In which of the
equilibrium systems
is HCO3− acting as a Brönsted-Lowry base?
− following
−
→
C.
HCO
+
H
S
H
CO
+
HS
A. HCO
H 2 O 3 − + HS
2−
2 − ←
2
3
→
B.
← H 2S ++ CO3 2−
3
− −− →
+ → 2−
D.
HCO
O +→
A.
CO3H 3O + CO3 −
B. HCO
MnO + HH
C. HCO3−334− ←
+ H22S ←
H 2 CO32−+ HS
←
− →
B.
+
HS
H
S
+ CO3
C. HCO
CH 33CHO
←
2
2−
D. HCO−3− + H 2 O→ →
H 3O + + CO
←
− 3
C.
HCO
+
H
S
H
CO
+
HS
D. CH 33COOH
← 2 3
2
−
→
D. conjugate
HCO3 + Hacid
+ CO32−
The
of H32O
O+ is
2O ←
-62−
A
spontaneous
redox
reaction
occurs
when
a
piece
of
iron
is
placed
in
1.0
M
CuSO
The conjugate
acid of H 2 O is
4.
A.
O
The reducing
agent is
−
B.
OH
The conjugate
acid of H 2 O is
2−
A. O +
C.
A. HFe
3O−
2−
A. O
B.
OH2+
D.
H
B. OH
Cu
− 2
2O
B.
C.
H 3O +
+
C. HH3O
O
C.
D.
H 22O2−
2
D.
D. HSO
2 O42
The strongest acid that can exist in an aqueous solution is
23.
The
strongest
acid
that
exist
an aqueous
solution
is
NH 2− acid
23. A.
The strongest
that
cancan
exist
in aninaqueous
solution
is
39. Consider +the following redox reaction:
B. H 3O− −
A. NH
NH2 2
C. HNO
3SO2 + 3H 2 O + ClO3− → 3SO 4 2− + 6H + + Cl −
2
B. HH33OO+ +
D. HClO 4
C.
C.
HNO2 2 half-reaction is
The HNO
reduction
D. HClO
D.
HClO4 4
A. ClO3− + 6H + → Cl − + 3H 2 O + 6e −
24. Which of the
following is possible for an acid?
B. ClO3− + 6H + + 6e − → Cl − + 3H 2 O
24. Which of the following is possible for an acid?
2−
+ 2H
+ 4H +for
+ 2e
C. SO
24. Which
of2 the
following
is4 possible
an− acid?
2 O → SO
ACID STRENGTH
CONCENTRATION
D. SO + 2H O + 2e − → SO 2− + 4H +
pH
B.
C.
D.
H 3O
HNO2
HClO 4
21 24. Which of the following is possible for an acid?
ACID STRENGTH
CONCENTRATION
pH
strong
0.01 M
2.0
A.
B.
weak
0.01 M
1.0
25. Consider the following equilibrium:
C.
strong
3M
5.5
+
−
25. Consider the following 2H
equilibrium:
→
2 O( l ) ←
D.
weak
3 MH 3O ( aq ) + OH ( aq )–0.5
→ H 3O + + OH −
2H
2 O( l ) to←
When comparing the
( aqequilibrium
)
( aq ) is reestablished.
A small amount of HCl is
added
water and
22 25. Consider
the
following
equilibrium:
new equilibrium with the original equilibrium,
A small amount of HCl is added to water
and equilibrium is reestablished. When comparing the
→
O +( aq ) + OH −( aq )
2Hthe
+
2 O( loriginal
) ← H 3equilibrium,
new
equilibrium
with
A. H 3O and pH both decreased.
[ ]
A
small amount of HCl
is added
to water and equilibrium is reestablished. When comparing the
pH
B.
and
pH both
both increased.
decreased.
A. [[HH OO ]] and
new equilibrium
with the original equilibrium,
andincreased.
pH decreased.
C.
and pH both
B. [[HH OO ]] increased
A. [ H O ] and pH both decreased.
and pH increased.
D.
increased
C. [H[HHOOO and
pH both increased. decreased.
B.
[ ] ]]decreased
and
pH increased.
D. [ H[ HO O] increased
and pH
decreased.
C.
] decreased
-733
3
3
33+
33+
++
++
++
OVER
+ +
3 3
D.
[ H O ]+decreased and pH increased.
[
[
3
+
]
]
26. The H 3O in 100.0 mL of 0.015 M KOH is
23 in 100.0 mL of 0.015 M KOH is
26. The H 3O + −13
A.
6.
7
+× 10
26. The [ H 3O ] in 100.0 mL of 0.015 M KOH is
B. 6. 7 × 10 −12
−13
A. 6. 7 × 10
−13−3
A.
6.
7
×
10
C. 1. 5 × 10 −12
B. 6. 7 × 10
B. 6. 7 × 10 −12−2
D. 1. 5 × 10
−3
−3
C. 1.1.
× 10
C.
5 ×510
−2 −2
D. 1.1.
× 10
D.
5 ×510
24 27. At any temperature, pK w is defined as
27. At any temperature, pK is defined as
w
27. A.
At any
pK is defined as
pKtemperature,
w = pH + pOH w
A.
pK w = pH + pOH
B.
pK w = pH − pOH
B.
A. pK
pKww == pH
pH−+pOH
pOH
C.
B. pK
pK w == pH
pH×−pOH
pOH
pK w w=
pH × pOH
pH
C. pK
pKw=w ==pHpH × pOH
D.
pK
D.
w
pOH
pOH
pH
D. pK w =
pOH
C.
28. The [ OH − ]− of a solution with pH 5.75 is
28.
The OH of a solution with pH 5.75 is
28.
[ ]
M a solution with pH 5.75 is
A.
× 10 of
The5.[6OH
]
M
A.
5.
6
×
10
B. 1.8 × 10 M
− −9
−6
−9
−6
−1 −9 M
B. 7.1.8
10
C.
6 ×6×10
M
A. 5.
× 10M
−1 −1
C. 9.7.2 6× ×
D.
1010
MM
−6
B. 1.8 × 10
−1
M
C.
pK w = pH × pOH
D.
pK w =
pH
pOH
25 28. The OH − of a solution with pH 5.75 is
[
]
A. 5. 6 × 10 −9 M
29. The value of−6K b for HPO 42− is
B. 1.8 × 10 M
−1
for HPO 42− is
29. The
−13bM
C.
7.
10
A. value
2.62 ××of
10K
−1
9.
29. D.
The
of
K Mfor HPO 42− is
B. value
6.22 ××10
10 −8
A. 2. 2 × 10 −13b
C. 1. 6 × 10 −7
−8
2−
26 B.
6.
2
×
10
−13
29. The
A. value
2. 2 ×of
10 K b for HPO 4 is
D. 4. 5 × 10−7−2
C.
−8
B. 1.6.62××10
10−13
−2
A.
2.
2
×
10
D.
× 10
10 −7
C. 4.
1. 65 ×
B. 6. 2 × 10 −8
D. 4. of
5 × 10−7−2
30. C.
Which
-81. 6 × the
10 following 0.10 M solutions is basic?
D.
5 ×the
10 −2
30. Which
of
following 0.10 M solutions
is basic?
A. 4.
LiCl
27 B. K 3of
PO 4 following 0.10 M solutions is basic?
30. A.
Which
LiCl the
C. NaClO
B. K 3PO 4 4
A. NH
LiCl
30. Which
of4the
0.10 M solutions is basic?
D.
NOfollowing
3
C.
NaClO
4
B. K 3PO 4
A. NH
LiCl4 NO3
D.
C. NaClO
B. K 3PO 4 4
D. NH 4the
NO3
31. C.
Consider
equilibrium for the indicator HInd at its transition point:
NaClO 4 following
28 D.
NH
NO
−
+
3
31. Consider4the following
equilibrium
the indicator
HInd
at its transition point:
HInd + H 2 Ofor→
← Ind + H 3O
−
+
31. Consider the following HInd
equilibrium
for→theInd
indicator
at its transition point:
+ H 3OHInd
2 O ←the
When a small amount of base+isHadded,
equilibrium
shifts to the
−
+
31. Consider the following equilibrium
for→theInd
indicator
HInd
at its transition point:
HInd
+is−Hadded,
+ H 3O
2O ←
When
a small
amount
of
base
the
equilibrium
shifts to the
>
Ind
A. left
and the
HInd
[
]
−
→ equilibrium
+−−Hadded,
+ H 3O + shifts to the
When
a small
amount
2 O ←theInd
B. left
left
and the
the
HIndof
Indis
[[ HInd
]]HInd
><base
Ind
A.
and
[ ]
[ ]
When
a small
amount
of
is ]added,
C.
right
andthe
the[[HInd
>
[ HInd
B.
and
]] <>]base
Ind
A. left
left
and
the
HInd
[ Ind
] the equilibrium shifts to the
[ Ind
D.
right
andthe
the[HInd
HInd
Ind ]
C.
and
the
[[HInd
B. right
left and
and
the
HInd
Ind
] ]><] ><[ Ind
[ Ind
A.
left
]
D.
and
the
[[HInd
C. right
rightand
andthe
the[ HInd
HInd
Ind] ]
[ Ind
B.
left
] ]<] <>[ Ind
D. right
right and
and the
the [HInd
HInd ] >< [Ind
Ind
C.
The approximate
K value
for the] indicator thymolphthalein is
D. right and the [ HInd ] < [ Ind ]
The
K value for the indicator thymolphthalein is
A. approximate
1 × 10
−− −
−
−− −
− −−
29 32.
32.
−−
a
−
−10
a
B. approximate
1 × 10 −4
32. A.
The
K a value for the indicator thymolphthalein is
1 × 10 −10
C. 4 −4
B. approximate
1 × 10
32. The
K a value for the indicator thymolphthalein is
A. 110× 10 −10
D.
C. 4 −4
B. 1 × 10−10
D.
A. 10
1 × 10
C. 4
B. 1 × 10 −4
D. 10
C. 4
D. 10
Acid Base Practice Test 31 HC
6. Nicotinic acid,
weakHC
acid
found in vitamin B.
6.6 HNicotinic
acid,
4 NO 2 , is a
6 H 4 NO 2 , is a weak acid found in vitamin B.
−5
Calculate the pH of 0.010
M HCthe
K a =M1. HC
4 × 610
. 2 K a = 1. 4 × 10 −5 . (4 marks)
Calculate
of2 0.010
H 4 NO
4 NO
6 HpH
(
32 ) (
)
What is the pH of a 0.150M solution of Na2SO4 (4 marks) (4 marks)
7. isAused
solution
of NaOHseparate
is used solutions
to neutralize
separate
solutions of HF and HBr.
7. A solution of NaOH
to neutralize
of HF
and HBr.
a) Write for
thethe
formula
equation of
forHF.
the neutralization of HF.
a) Write the formula equation
neutralization
(1 mark)
(1 mark)
(1 mark)
(1 mark)
b) Write the net ionic equation for the neutralization of HBr.
b) Write the net ionic
equation for the neutralization of HBr.
c) One of the neutralization reactions above produces a salt that undergoes hydrolysis.
c) One of the
neutralization reactions above produces a salt that undergoes hydrolysis.
Identify
the ionic
salt and
write the
equation
for the hydrolysis
reaction.
(2 marks)
Identify the salt and write
the net
equation
for net
the ionic
hydrolysis
reaction.
(2 marks)