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What is pH of 0.0002 M HI solution? What
will be pH if 1 mL of the above was added
to 999 mL of pure water?
What is pH of mixture prepared by mixing
20 mL 0.07 M NaOH and 13 mL 0.09 M
HCl?
Calculate pH of 1 M solution of nitrous
acid, pKa=3.37.
Calculate pH of 0.001 M solution of
benzoic acid, pKa=4.19.
0.1 M solution of weak acid has pH=4.0.
Calculate pKa.
20 mL of 0.1 M solution of weak acid was
mixed with 8 mL 0.1 M solution of NaOH.
Measured pH was 5.12. Calculate pKa.
Calculate the pH of a 0.2M solution of
ammonia. Ka=5.62×10-10.
Calculate pH of 0.01 M aniline
hydrochloride. Aniline pKb=9.4.
Calculate pH of 0.1 M potassium
hydrogen oxalate. pKa1=1.25, pKa2=4.27.
Calculate pH of solution prepared from
0.1 mole of formic acid and 0.02 mole of
NaOH diluted to 1L. PKa=3.75.
You have 100 mL of 1 M ammonia
solution (pKa=9.25). What volume of 1 M
hydrochloric acid is needed to prepare
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What is pH of 0.0002M HI solution?
What will be pH if 1 mL of the above was
added to 999 mL of pure water?
HI is a strong acid so we can assume it is fully
dissociated.
3×10-4 M solution is concentrated enough to allow
us to neglect water dissociation (see table in
strong acid section). Thus we can use 7.6 equation
7.6
and the pH calculated is -log(3×10-4)=3.70.
After the solution has been diluted 1000 times HI
concentration is 2×10-7 M and we are in the area
where water dissociation may spoil results of the
7.6 equation. Thus it wil be better to use the
equation 7.4
7.4
and calculated pH=6.62 (you know how to solve
quadratic equation, don't you?).
Using equation 7.6 we will get pH=6.70. Difference
is not large - but clearly visible.
As additional confirmation our pH calculator shows
exactly the same results - 3.70 and 6.62.
2
What is pH of mixture prepared by mixing
20 mL 0.07 M NaOH and 13 mL 0.09 M HCl?
Although you are asked about pH, this is a simple
limiting reagent question.
When you mix strong acid and strong base they
react till one of the reagents runs out. What is left
is responsible for pH.
Reaction taking place in the solution is
NaOH + HCl -> NaCl + H2O
Initially there were 20×0.07=1.4 mmol of NaOH
and 13×0.09=1.17 mmol of HCl. They react 1:1,
so after reaction 1.4-1.17=0.23 mmol of NaOH is
left. Final volume of the solution is 20+13=33 mL,
thus final concentration of NaOH left is
0.23/33=0.0070 M. pOH=-log(0.0070)=2.15 and
pH=14-2.15=11.85.
Note that if there is exactly the same amount of
both reactants reaction proceeds to the end - and
you are left with perfectly neutral solution of
pH=7.00.
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Calculate pH of 1M solution of nitrous acid,
pKa=3.37.
Let's try first with the simplified formula 8.13.
8.13
If we put concentration and Ka values into formula
we get [H+]=0.021 M and pH=1.69. Is it correct?
To be sure we have to check if the 5% rule is
obeyed, that is if the dissociation fraction is less
then 5%.
1.3
Dissociated acid concentration is the same as [H+]
- if we assume we can neglect water dissociation.
It is obvious that we can, as 0.021 M is five orders
of magnitude larger than the concentration of H+
from the water. So dissociation fraction is 0.021/1
- a little bit over 2%, much less then 5%, so the
5% rule is obeyed.
As additional confirmation our pH calculat
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Calculate pH of 0.001M solution of benzoic
acid, pKa=4.19.
We will start with the simplified formula 8.13.
8.13
From formula we get [H+]=2.54×10-4 M and
pH=3.60. Is it correct? To be sure we have to
check if the 5% rule is obeyed, that is if the
dissociation fraction is less then 5%.
1.3
Dissociated acid concentration is the same as [H+],
thus dissociation fraction is 2.54×10-4/0.001 above 25%. We can't use simplified formula 8.13!
We have to use more precise equation 8.10:
8.10
This time we get pH=3.65, exactly the
0.1 M solution of weak acid has pH=4.0.
Calculate pKa.
If pH=4.0 then [H+]=10-4 - that means that
dissociation fraction is about
10-4/0.1×100%=0.1% - well below 5%. That in
turn means that pH of the solution is described by
the simplified formula 8.13.
8.13
All we have to do is rearrange to get equation for
Ka:
5
and Ka=(10-4)2/0.1=10-7 and pKa=7.
Our pH calculator can help us check this result and indeed, entering 0.1 as concentration and 7 as
pKa we get pH=4.0 which confirms our
calculations.
20 mL of 0.1 M solution of weak acid was
mixed with 8 mL 0.1 M solution of NaOH.
Measured pH was 5.12. Calculate pKa.
In the case of not-so-weak acids you can assume
the neutralization reaction is quantitative (all of
the strong base reacts with the weak acid) and the
pH of the solution is described by HendersonHasselbalch equation:
15.2
All you have to do is to find out concentrations of
A- and HA. However, it is worth of noting here,
that you can save part of the calculations,
replacing concentrations with numbers of moles volume is the same for both substances so it
cancels out.
Solution was prepared using 20×0.1=2 mmol of
acid and 8×0.1=0.8 mmol of strong base. If the
reaction was quantitative after the neutralization
there was 1.2 mmol of acid HA left and 0.8 mmol
of A-. Inserting these values into HendersonHasselbalch equation gives
5.12=pKa+log(0.8/1.2). Solving for pKa you get
pKa=5.12-log(0.8/1.2)=5.12+0.18=5.30.
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pH calculator can help you check this result. Enter
5.3 as pKa, 0.1 as concentration of acid, select
sodium hydroxide as a base, enter 0.1 as its
concentration, finally select Ca+Cb+Va+Vb as a
way of expressing solution composition and enter
20 and 8 as acid and base solutions volumes - and
the pH displayed will be 5.12, confirming the
result.
Note that for strong or very weak acids and/or for
very diluted solutions this method may give wrong
results - when pH is further changed by
dissociation or hydrolyzis or when the water
autodissociation can't be neglected.
Calculate the pH of a 0.2M solution of
ammonia. Ka=5.62×10-10.
Ammonia is a weak base, so the most convenient
approach is to calculate pOH using Kb, and then to
convert it to pH.
From the Brønsted-Lowry theory we know that
Ka×Kb=Kw, that allows us to calculate
Kb=Kw/Ka=10-14/5.62×10-10=1.78×10-5.
Calculation of the pOH will be now identical with
the calculation of pH of a weak acid.
Let's try first with the simplified formula 8.17:
8.17
If we put concentration and Kb values into formula
we get [OH+]=0.00189 M and pOH=2.72. Is it
correct? To be sure we have to check if the 5%
rule is obeyed, that is if the dissociation fraction is
less then 5%.
1.3
7
Dissociated base concentration is the same as [OH] - if we assume that we can neglect water
dissociation. It is obvious that we can, as
0.00189 M is about five orders of magnitude larger
than the concentration of OH- from the water. So
dissociation fraction is 0.00189/0.2 - below 1%,
much less then 5%, so the 5% rule is obeyed.
Now we need the final touch - we know pOH=2.72,
but we need pH. Time to use equation 1.6.
pH=14-2.72=11.28.
Our pH calculator shows 11.27. Difference in the
last digit can be probably attributed to different
dissociation constant used.
Calculate pH of 0.01M aniline hydrochloride.
Aniline pKb=9.4.
Aniline in solution of its hydrochloride is in form of
conjugate acid. To calculate pH of such solution we
have to find pKa for aniline and treat it like weak
acid.
Knowing pKb we can find pKa from the equation
2.5:
2.5
pKa=14-9.4=4.6 - and this value we will put into
simplified formula 8.13.
8.13
Result is [H+]=5.0×10-4 M and pH=3.30. We have
used simplified formula, so now we have to check
if the 5% rule is obeyed:
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1.3
Dissociation fraction is 5.0×10-4/0.01 - exactly
5%. We were allowed to use equation 8.13.
Our pH calculator doesn't calculate pH for salts,
but salt is nothing else but equimolar solution of
acid and base. BATE calculates pH as 3.31 - where
is the source of difference?
First suspect is 5% rule - it doesn't state results
will be exact, but the error will be acceptable. Let's
try to calculate the result using more precise
formula 8.10.
8.10
This time [H+]=4.89×10-4 and pH=3.31. This
result is more precise - but is it better and worth
additional calculations?
No!
Dissociation constant is given only with two
significant digits accuracy, so our pH is in fact 3.3
- or even just 3, as the concentration is given with
only one significant digit.
pH calculator always calculate the result with full
precision - that's what computers are for. It is up
to us - humans - to understand the results.
Calculate pH of 0.1M potassium hydrogen
oxalate. pKa1=1.25, pKa2=4.27.
This question - if done by hand - requires either
approximate method (that gives questionable
results) or tedious calculations.
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In amphiprotic salt section we have derived
equation 12.9
12.9
According to the analysis we have done (results
shown in the table) equation 12.9 gives good
results for more concentrated solutions. 0.01 M is
a little bit too diluted for the precise result.
PH=1/2(1.25+4.27
) =2.76.
Let's check this result using pH calculator. To
create potassium hydrogen oxalate solution you
have to select oxalic acid and potassium hydroxide
and to enter 0.1 as both concentrations.
pH calculator shows 2.86 - and this is the correct
result. Approximate method is very fast - but gives
only very rough results.
Calculate pH of solution prepared from
0.1 mole of formic acid and 0.02 mole of
NaOH diluted to 1L. pKa=3.75.
This is classic buffer question that we will solve
using Henderson-Hasselbalch equation.
15.2
We need to know concentration of HCOO- and
HCOOH. Formic acid has pKa value larger then 2.5
(see pH of buffer section) so we can assume
stoichiometry is solely responsible for
concentrations of both acid and conjugate base. If
so, after neutralization has taken place
[HCOOH]=0.1-0.02=0.08 M and [HCOO-]=0.02 M.
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Putting these values into Henderson-Hasselbalch
equation we get:
pH=3.75+log(0.02/0.08)=3.15.
Using pH calculator we get 3.17. Our rule of thumb (the one about pKa>=2.5)
states that difference between real pH and calculated pH will be acceptable, not zero.
You have 100 mL of 1 M ammonia solution
(pKa=9.25). What volume of 1 M hydrochloric
acid is needed to prepare buffer with
pH=9.5?
Adding hydrochloric acid to the solution of
ammonia (base) we create a conjugate acid NH4+.
Ratio of their concentrations at pH 9.5 will be
given by the Henderson-Hasselbalch equation.
15.2
If all added acid reacts with ammonia, amounts of
conjugate acid NH4+ created and base NH3 left is
defined by the stoichiometry of the neutralization
reaction:
qbq2.1
and
qbq2.2
qbq2.3
qbq2.3 is equivalent to
qbq2.4
where na and nb are number of moles of acid and
conjugate base, nNH3start is number of moles of
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ammonia at the beginning (0.1 mole - n=CV) and
nHCl is number of moles of acid added. We are
using number of moles as it makes calculation
much easier - this way we don't have to calculate
how the concentrations change due to the dilution
once the acid is added to original ammonia
solution. We can do that thanks to the fact that in
the Henderson-Hasselbalch equation volumes can
be canceled out and replaced by mole ratio.
Putting pH=9.5 into the Henderson-Hasselbalch
equation we get
qbq2.5
Finally we have a set of equations qbq2.4 and
qbq2.5. When solved they give na=0.036 and
nb=0.064. From qbq2.2 we get that we have to
add 0.036 mole of HCl, or 36 mL of 1 M solution.
We can use pH calculator to check the result.
Create a solution containing both ammonia and
hydrochloric acid. Enter 1 as both concentrations.
From the drop down list above pH sign select
Ca+Cb+Va+Vb. Enter 36 as acid volume and 100
as base volume. Perfect!
Note: if you need program that will help in buffer
calculation, our pH calculator is not suited for the
task, however, you can try our Buffer Maker - the
buffer calculator. Buffer calculator was specifically
designed to help in buffer preparation, so - given
pH - it will automatically calculate recipe of any
buffer you may need.
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