Acids and Bases 2  Questions
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Acids
and Bases (strong acids and bases; conjugate acid/base pairs;
pH;selfdissociation
of water)
Weak
Acids and Bases (weak acids; weak bases; K_{a}
and K_{b})
Calculations
with Weak Acids and Bases (calculations involving weak acids
and bases and buffers)
Acidbase
Titrations (titration curves for acid/base titrations; indicators)

Shortcut to Questions
Q: 1 2 3
4 5
6 7 8 9
10 11

1

Write an equation for the acid/base reaction in water solution between equimolar amounts of each of the following pairs and, using ionization constants, predict in each case whether reactants or products predominate at equilibrium
(1.) HSO_{4}^{}, SO_{3}^{2}
(2.) H_{2}S, H_{2}BO_{3}^{}
(3.) CH_{3}COOH, ClCH_{2}CO_{2}^{}
(4.) CO_{3}^{2}, H_{2}CO_{3}
(5.) NH_{4}^{+}, NO_{2}^{}
(6.) C_{6}H_{5}OH, HCO_{3}^{}
(7.) C_{6}H_{5}COOH, HCO_{3}^{}
(8.) Fe(OH_{2})_{6}^{3+}, F^{}
(9.) N_{2}H_{5}^{+}, CN^{}
(10.) HCO_{3}^{}, HPO_{4}^{2}


2 
Express the following H_{3}O^{+} concentrations in terms of pH:
(1.) 1.0 x 10^{3} M
(2.) 5.4 x 10^{9} M


3 
Calculate [H_{3}O^{+}] for a solution of pH 8.37.


4 
Calculate the pH of a 0.10 M water solution of propanoic acid (CH_{3}CH_{2}COOH)


5 
Calculate the pH of a 0.040 M aqueous solution of hydrogen cyanide


6 
Calculate the pH of a
0.50 M water solution of ethylamine (CH_{3}CH_{2}NH_{2})


7 
At 273 K the ionic product constant of water (K_{w})
is 1.15 x 10^{15} M^{2}.
(1.) Calculate the hydrogen ion concentration of a neutral solution at that temperature.
(2.) Calculate pOH and [OH^{}] for a solution of pH 4.07 at 273 K.


8 
A particular water solution of formic acid is 3.2% ionised at 25 °
C. What is the pH of the solution?


9 
What is the pH of a 1.0
M solution of ethanol in water?


10 
Sodium (0.010 mole) is
added to water (2.0 litre)
(1.) What is the pH of the resulting solution?
(2.) What mass of gas is given off in the reaction?


11 
The average pH of beer is 4.40. If the acidity is regarded as due only to the presence of dissolved carbon dioxide from the fermentation process, calculate the concentration of the carbonic acid in the beer. (Acidity arising from the second dissociation of carbonic acid or from the dissociation of water is to be neglected).


Acids
and Bases 2 (Answers) 

1

The fundamental law of
acid/base reactions states: "the stronger acid and base combine
to give the weaker conjugates." This applies even when both
acids involved are weak, but instead of complete reaction occurring
as would be the case if strong species were involved, an equilibrium
mixture results and the equilibrium lies towards the side with the
weaker acid and base. All the parts in Q1 involve weak species and
hence the use of equilibrium arrows in the following equations.
(1.)
HSO_{4}^{} + SO_{3}^{2}
SO_{4}^{2} + HSO_{3}^{}
K_{A} of HSO_{4}^{}
= 10^{1.99} > K_{A} of HSO3^{}
= 10^{7.20}
Therefore HSO_{4}^{} is the
stronger acid and products will predominate.
(2.) H_{2}S
+ H_{2}BO_{3}^{}
HS^{} + H_{3}BO_{3}
K_{A} of H_{2}S = 10^{7.02}
> K_{A} of H_{3}BO_{3}= 10^{9.24}
Therefore H_{2}S is the stronger acid
and products will predominate
(3.)
CH_{3}COOH + ClCH_{2}CO_{2}^{}
CH3CO_{2}^{} + ClCH_{2}COOH
K_{A} of ClCH_{2}COOH
= 10^{2.86} > K_{A} of CH_{3}COOH
= 10^{4.76}
Therefore ClCH_{2}COOH is the stronger
acid and reactants will predominate
(4.)
CO_{3}^{2} + H_{2}CO_{3}
HCO_{3}^{} + HCO_{3}
^{}
K_{A} of H_{2}CO_{3}
= 10^{6.55} > K_{A} of HCO_{3}
^{} = 10^{10.33}
Therefore H_{2}CO_{3} is the
stronger acid and reactants will predominate.
(Note that there was no need to consult K_{A
}data as carbonic acid is a diprotic weak acid and its
conjugate base, hydrogencarbonate ion, must necessarily be weaker
because it has the same structure but carries a negative charge,
making it more difficult for a H^{+} ion to be released.)
(5.)
NH_{4}^{+} + NO_{2}^{}
NH3 + HNO_{2}
K_{A} of HNO_{2} = 10^{3.14}
> K_{A} of NH_{3} =
10^{9.24}
Therefore HNO_{2} is the stronger acid
and reactants will predominate.
(6.)C_{6}H_{5}OH
+ HCO_{3}^{}
C_{6}H_{5}O^{} + H_{2}CO_{3}
K_{A} of H_{2}CO_{3}=
10^{6.55} > K_{A}
of C_{6}H_{5}OH =
10^{9.98}
Therefore H_{2}CO_{3}
is the stronger acid and reactants will predominate.
(7.) C_{6}H_{5}COOH
+ HCO_{3}^{}
C_{6}H_{5}CO_{2}^{}
+ H_{2}CO_{3}
K_{A}
of C_{6}H_{5}COOH
= 10^{4.20}
> K_{A} of H_{2}CO_{3}
= 10^{6.55}
Therefore C_{6}H_{5}COOH
is the stronger acid and products will predominate.
(8.)
Fe(OH_{2})_{6}^{3+} + F^{}
Fe(OH_{2})_{5}(OH)^{2+}
+ HF
K_{A} of Fe(OH_{2})_{6}^{3+}
= 10^{2.17} > K_{A} of
HF = 10^{3.17}
Therefore Fe(OH_{2})_{6}^{3+}
is the stronger acid and products will predominate.
(9.)
N_{2}H_{5}^{+} + CN^{}
N_{2}H_{4}+
HCN
K_{A} of N_{2}H_{5}^{+}
= 10^{7.94}> K_{A} of
HCN = 10^{9.22}
Therefore N_{2}H_{5}^{+}
is the stronger acid and products will predominate.
(10.)
HCO_{3}^{} + HPO_{4}^{2}
H2CO_{3} + H_{2}PO_{4}^{}
In this case, all four species ae potentially acids but both the
species on the products side are diprotic acids and their
conjugate bases are the reactants, so as electrically
neutral H2CO_{3} must be a stronger acid than its negatively
charged conjugate, HCO_{3}^{}
, and single negatively
charged H_{2}PO_{4}^{} must be a stronger
acid than its double negatively charged conjugate, HPO_{4}^{2}
, then the equilibrium lies to the reactants side.


2 
pH = log [H_{3}O^{+}]
(1.) pH = log (1.0 x 10^{3} M) = 3.00
(2.) pH = log (5.4 x 10^{9} M) = 8.27
Note the conversion of significant figures between concentration and pH  the number of significant figures in the concentration becomes the number of decimal places in the pH value (and vice versa), since pH is a logarithmic scale. So in the examples above, 2 significant figures of accuracy in the concentration data results in 2 decimal places of accuracy in the resultant pH values.
It is usual practice to restrict the accuracy of pH values to a maximum of 2 decimal places, since measurement approximations used to calculate pH limit it's inherent accuracy.


3 
[H_{3}O^{+}] = 10^{(pH)}.
i.e. hydrogen ion concentration is found from the inverse log of
the negative pH value.
\
Given a pH of 8.37, [H_{3}O^{+}] = 10^{8.37} = 4.3 x 10^{9} M.


4

To calculate pH we need
to know [H_{3}O^{+}]. For a weak acid such as propanoic
acid, which does not dissociate completely in water, [H_{3}O^{+}]
at equilibrium does not equal the initial concentration of the acid.
So [H_{3}O^{+}] must first be calculated:
pK_{A}
of propanoic acid = 4.87 (from data sheet).
K_{A}
= 10^{ 4.87}
Equilibrium equation:
CH_{3}CH_{2}COOH + H_{2}O
CH_{3}CH_{2}CO_{2}^{} + H_{3}O^{+}.
From this equation we
construct the expression for K_{A}:
K_{A} = [CH_{3}CH_{2}CO_{2}^{}]
[H_{3}O^{+}] / [CH_{3}CH_{2}COOH]
It is apparent from the
equilibrium equation that each acid molecule that dissociates has
to form one H_{3}O^{+} and one CH_{3}CH_{2}CO_{2}^{}
ion.
Therefore [H_{3}O^{+}]
= [CH_{3}CH_{2}CO_{2}^{}] (i.e. their concentrations
must be the same).
The remaining variable
is [CH_{3}CH_{2}COOH]. Since propanoic acid is weak
and only a very small amount is ionised in solution, we make the
assumption that the equilibrium concentration of acid equals the
initial concentration of acid (this allows us to avoid creating
a quadratic expression, which is relatively time consuming to solve.)
So in this example, we assume that the equilibrium concentration
of propanoic acid = 0.10 M.
Note that the validity
of this assumption can be tested after calculation but for the purposes
of exam questions in this chemistry course you can assume that the
assumption is always valid. The worked problems presented in most
textbooks do not assume the assumption is valid, and hence include
checks as part of the working.
Rearranging the expression
for K_{A}:
[H_{3}O^{+}]
[CH_{3}CH_{2}CO_{2}^{}] = K_{A}
[CH_{3}CH_{2}COOH]
but since [H_{3}O^{+}]
= [CH_{3}CH_{2}CO_{2}^{}],
[H_{3}O^{+}]^{2}
= K_{A} [CH_{3}CH_{2}COOH]
Substituting values we
obtain
[H_{3}O^{+}]^{2}
= 10^{ 4.87} x 0.10 = 10^{5.87} M^{2}
^{ }
\
[H_{3}O^{+}] = 10^{5.87/2} M = 10^{2.94}
M = 1.2 x 10^{3} M
Finally, from [H_{3}O^{+}]
we obtain the pH:
pH = log [H_{3}O^{+}]
= log (1.2 x 10^{3} M) = 2.92.


5 
HCN + H_{2}O
H_{3}O^{+} + CN^{} ^{
}
We find pH from [H_{3}O^{+}],
which is in turn calculated from K_{A} for HCN
pK_{A}
of HCN = 9.22 (from data sheet).
K_{A}
= 10^{ 9.22}
^{
}
K_{A} = [H_{3}O^{+}]
[CN^{}] / [HCN]
Since [H_{3}O^{+}]
= [CN^{}]
\ K_{A}
= [H_{3}O^{+}]^{2} / [HCN]
We assume [HCN] at equilibrium
is equal to initial [HCN] as the amount of HCN that has dissociated
is negligible compared with its initial concentration.
Substituting and rearranging:
K_{A}
= 10^{ 9.22} = [H_{3}O^{+}]^{2}
/ 0.040 M
[H_{3}O^{+}]^{2}
= 10^{ 9.22} x 0.040 = 10^{ 10.62 }M^{2}
^{ }
\
[H_{3}O^{+}] = 10^{10.62/2 } M = ^{
}10^{ 5.31 }M
pH = log [H_{3}O^{+}]
= log ( 10^{ 5.31 }) = 5.31.


6 
Ethylamine, CH_{3}CH_{2}NH_{2}, has the
same structure as the weak base, ammonia, NH_{3}, except
that one of the H atoms has been replaced by an ethyl group, CH_{3}CH_{2},
but it is still a base as the central N atom has a lone pair of
electrons that can bond to an H^{+} donated by an acid.
The equation for the equilibrium set up by ethylamine and water
is:
CH_{3}CH_{2}NH_{2} + H_{2}O
CH_{3}CH_{2}NH_{3}^{+} + OH^{}
pK_{A} of the conjugate acid of ethylamine, the
ethylammonium ion, CH_{3}CH_{2}NH_{3}^{+}
= 10.67.
Therefore pK_{B} of CH_{3}CH_{2}NH_{2}
(the conjugate base of CH_{3}CH_{2}NH_{3}^{+})
= pK_{W}  pK_{A}(CH_{3}CH_{2}NH_{3}^{+})
= 14  10.67 = 3.33.
\
K_{B} for ethylamine= 10^{ 3.33} M.
K_{B} = [CH_{3}CH_{2}NH_{3}^{+}]
[OH^{}] / [CH_{3}CH_{2}NH_{2}]
As there is no other source of hydroxide ions, [CH_{3}CH_{2}NH_{3}^{+}]
= [OH^{}]
\
K_{B} = [OH^{}]^{2} / [CH_{3}CH_{2}NH_{2}]
We can take [CH_{3}CH_{2}NH_{2}] at
equilibrium as equal to initial [CH_{3}CH_{2}NH_{2}]
as the extent of hydrolysis of a weak base is negligible compared
with its initial concentration.
10^{ 3.33 } = [OH^{}]^{2} / 0.50 M
[OH^{}]^{2} = 10^{ 3.33} x 0.50 = 10^{
3.63} M^{2}
[OH^{}] = 10^{ 3.63/2 } = 10^{ 1.82 }M
pOH = log [OH^{}] = 1.82
\ pH = 14  pOH = 14  1.82 = 12.18.


7 
(1.)
Equation for the selfionization of water:
H_{2}O + H_{2}O
H_{3}O^{+} + OH^{}
By definition, K_{W}
= [H_{3}O^{+}] [OH^{}].
Since [H_{3}O^{+}] = [OH^{}], K_{W} = [H_{3}O^{+}]^{2}
\
[H_{3}O^{+}] = (K_{W})^{½} = (1.15 x 10^{15})^{½} = 3.39 x 10^{8} M
(2.) pH = 4.07.
[H^{+}] = 10^{(pH)} = 10^{4.07} = 8.51
x 10^{5} M
[OH^{}] = K_{W} / [H^{+}]
= 1.15 x 10^{15} / 8.51 x 10^{5} = 1.35 x 10^{11}
M
\
pOH = log [OH^{}] = log (1.35 x 10^{11} M) =
10.87.


8 
The first step is to
calculate the concentration of the HCOOH solution from K_{A}
and the ionisation information.
HCOOH + H_{2}O
HCO_{2}^{} + H_{3}O^{+}
pK_{A}
= 3.74 (from data sheet)
K_{A}
= 10 ^{3.74} M = 1.84 x 10^{4} M = [HCO_{2}^{}]
[H_{3}O^{+}] / [HCOOH]
"3.2 % ionised" implies
that there are 3.2 HCO_{2}^{}and 3.2 H_{3}O^{+}
ions per 100 HCOOH molecules initially present. Let the initial
[HCOOH] = Z M.
Then [HCO_{2}^{}]
= [H_{3}O^{+}] = 0.032 x Z M, and strictly the
equilibrium [HCOOH] =
Z  0.032 Z = Z(1  0.032) = 0.968 Z M.
However, following the
procedure used previously, the equilibrium [HCOOH] can be taken
as Z M.
Substituting this in
the expression for K_{A}
:
K_{A}
= [HCO_{2}^{}][H_{3}O^{+}]
/ [HCOOH]
= (0.032 x Z)(0.032 x
Z) / Z = 1.84 x 10^{4} M
\ (0.032
x Z)(0.032) = 1.84 x 10^{4} M
Z = [HCOOH] = 1.84 x
10^{4} / (0.032 x 0.032) = 0.18 M.
Now that the concentration
is known, the pH is easily determined:
K_{A} = [H_{3}O^{+}]^{2}
/ [HCOOH]
\
[H_{3}O^{+}] = (1.84 x 10^{4} x 0.18)^{½}
= 5.75 x 10^{3} M.
pH = log [H_{3}O^{+}]
= log (5.70 x 10^{3} M) = 2.24


9 
Ethanol acting as acid:
CH_{3}CH_{2}OH + H_{2}O CH_{3}CH_{2}O^{}
+ H_{3}O^{+} ^{
}
Water acting as acid:
H_{2}O + H_{2}O OH^{} + H_{3}O^{+} ^{
}
pK_{A}
of CH_{3}CH_{2}OH is greater than 14 (from data
sheet). This means that H_{2}O (which has a pK_{A}
equal to 14) is a stronger acid than ethanol. The contribution
of H_{3}O^{+} from the self ionisation of water
is greater than the contribution from ethanol, and the pH of the
solution is determined by the water present. As a result,
the pH of the ethanol solution is 7.00  the same as if the ethanol
were not present.


10 
Na + H_{2}O ®
NaOH + ½H_{2}
_{
}
(1.) From the above equation 0.010 mole sodium will react to form 0.010 mole NaOH. Molarity = no. mol / volume = 0.010 mol / 2.0 L = 0.0050 M. NaOH is very soluble, and completely dissociates in solution, so [OH^{}] = 0.0050 M.
[H_{3}O^{+}] = K_{W} / [OH^{}] = 1.0
x 10^{14} M^{2} / 0.0050 M
= 2.0 x 10^{12} M.
\pH = log (2.0 x 10^{12} M) = 11.7
(2.)
Mass of H_{2} gas = no. mol H_{2} x molar mass H_{2}
= 0.50 mol x 2.016 g mol^{1} = 1.0
g.


11 
pH = 4.40,
\[H_{3}O^{+}]
= 10^{4.40} M.
H_{2}CO_{3}
+ H_{2}O H_{3}O^{+} + HCO_{3}^{} ^{
}
pK_{A}
of H_{2}CO_{3} = 6.35.
K_{A} = [H_{3}O^{+}]
[HCO_{3}^{}] / [H_{2}CO_{3}] =
10^{6.35} M
[H_{3}O^{+}]
= [HCO_{3}^{}]
\
[H_{2}CO_{3}] = [H_{3}O^{+}]^{2}
/ K_{A}
= (3.98 x 10^{5}
M)^{2} / 4.47 x 10^{7} M
= 3.5 x 10^{3}
M.

