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Mix 569.9 ml of 0.200M
When the acid is added to the carbonate solution, the
This means we are interested in the 2nd dissociation of carbonic acid:
Rearranging:
This is the ratio of acid : cobase we need to achieve to get the target pH of 9.45.
A certain volume of HCl which I will call V must be added such that the total volume of buffer created is 1 Litre.
From the original equation you can see that if n moles of HCl is added, then n moles of
Since
So we can say that the no. moles
Since the total volume must be 1 Litre we can say that the initial moles of
So the no. moles of
Since the total volume is common, we can put these into
This is the volume of
The volume of
Iteration check:
So that's all good.
#["H"^(+)] = 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))#
Note that this works best if
DISCLAIMER: DERIVATION!
Well, concentration is a state function, so we can simply choose acid 1 to go first, and acid 2 can go second, suppressed by the equilibrium of the first.
By writing out an ICE table, you would construct the mass action expression for acid 1 (say,
#"HA"(aq) rightleftharpoons "H"^(+)(aq) + "A"^()(aq)#
#K_(a1) = (["H"^(+)]_1["A"^()])/(["HA"]) = (["H"^(+)]_1alpha_1C_1)/((1  alpha_1)C_1)#
Although we know that
Since we assume
#K_(a1) ~~ (["H"^(+)]_1alpha_1C_1)/(C_1) = alpha_1["H"^(+)]_1#
So,
#["H"^(+)]_1 ~~ K_(a1)/alpha_1#
The second acid, say
#"BH"^(+)(aq) rightleftharpoons "B"(aq) + "H"^(+)(aq)#
And the
#K_(a2) = (["B"]["H"^(+)])/(["BH"^(+)]) = (["H"^(+)]alpha_2C_2)/((1  alpha_2)C_2)#
And since we also have
#K_(a2) ~~ alpha_2["H"^(+)]#
#["H"^(+)] ~~ K_(a2)/alpha_2#
with
#alpha_2C_2 = ["H"^(+)]_2 = K_(a2)/alpha_2  alpha_1C_1# .
So, one form of this is
#["H"^(+)] = ["H"^(+)]_1 + ["H"^(+)]_2#
#= alpha_1C_1 + alpha_2C_2#
But we have built into
#color(red)(["H"^(+)] < (K_(a1))/alpha_1 + K_(a2)/alpha_2)#
It would be convenient to determine
#K_(a2)/alpha_2 = alpha_1C_1 + alpha_2C_2#
#0 = C_2alpha_2^2 + alpha_1C_1alpha_2  K_(a2)#
This becomes a quadratic equation. If you wish to see it,
#color(green)(alpha_2) = ((alpha_1C_1) pm sqrt(alpha_1^2C_1^2  4C_2(K_(a2))))/(2C_2)#
#= color(green)(((alpha_1C_1) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2)))/(2C_2))#
And so,
#["H"^(+)] = alpha_1C_1 + ( (alpha_1C_1)/(2C_2) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2))/(2C_2))C_2#
#= 1/2 alpha_1C_1 pm sqrt(((alpha_1C_1)/(2))^2 + K_(a2)C_2)#
And lastly, it would be convenient to know this in terms of
Since
#alpha_1 = sqrt(K_(a1)/C_1)#
Therefore:
#color(blue)(["H"^(+)]) = 1/2 sqrt(K_(a1)C_1) pm sqrt(((C_1)/(2))^2K_(a1)/C_1 + K_(a2)C_2)#
#= 1/2 sqrt(K_(a1)C_1) pm sqrt(1/4K_(a1)C_1 + K_(a2)C_2)#
#= color(blue)(1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2)))#
TESTING ON ACTUAL PROBLEM
And of course, we should try this on an actual problem.
Consider acetic acid (
#"HCHO"_2(aq) rightleftharpoons "H"^(+)(aq) + "CHO"_2^()(aq)#
#1.8 xx 10^(4) = x^2/(0.50  x) ~~ x^2/0.50#
So,
#["H"^(+)]_1 ~~ sqrt(1.8 xx 10^(4) cdot 0.50) = "0.00949 M"#
and
#alpha_1 ~~ x/(["HCHO"_2]) = 0.019#
Let's see if we get the same
#1.8 xx 10^(4) stackrel(?)(=) (alpha_1C_1)^2/((1  alpha_1)C_1)#
#= (0.019 cdot 0.50)^2/((1  0.019)cdot0.50) ~~ 1.8 xx 10^(4) color(blue)(sqrt"")#
Next, add in acetic acid.
#"HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O"_2^()(aq)#
#1.8 xx 10^(5) = ((["H"^(+)]_1 + x)(x))/(2.00  x) ~~ (["H"^(+)]_1 + x)x/2#
#~~ ["H"^(+)]_1x/2 + x^2/2#
The quadratic equation solves to be
#x = ["C"_2"H"_3"O"_2^()] = "0.00291 M"# ,which is also the
#"H"^(+)# contributed by the second acid.
And so, the total
#color(green)(["H"^(+)]) = "0.00949 M" + "0.00291 M" = color(green)ul("0.0124 M")#
This acid's percent dissociation at this concentration is
#alpha_2 = x/(["HC"_2"H"_3"O"_2]) = "0.00291 M"/"2.00 M" = 0.00145# .
So from the first form of the derived equation,
#color(green)(["H"^(+)] = alpha_1C_1 + alpha_2C_2)#
#= 0.019 cdot 0.50 + 0.00145 cdot 2.00 = color(green)ul("0.0124 M")# #color(blue)(sqrt"")#
Now to check the general formula we derived above.
#color(green)(["H"^(+)]) stackrel(?)(=) 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))#
#= 1/2 (sqrt(1.8 xx 10^(4) cdot 0.50) pm sqrt(1.8 xx 10^(4) cdot 0.50 + 4 cdot 1.8 xx 10^(5) cdot 2.00))#
#= 1/2(sqrt(9.00 xx 10^(5)) pm sqrt(9.00 xx 10^(5) + 1.44 xx 10^(4)))#
#=# #color(green)ul("0.0124 M")# #color(blue)(sqrt"")#
Here's what I get.
Point group of
It has a
The image below shows the three
(Adapted from SlidePlayer)
If you look at the molecule from the top, the
1.
Its structure is
Its only symmetry element is a single
Thus, it has lost two
2.
That now belongs to point group
Its structure is
Thus, it has lost even the
Here's what's going on here.
You're actually dealing with a disproportionation reaction here. In a disproportionation reaction, the same element undergoes both oxidation and reduction.
In this case, manganese(VI) is reduced to manganese(IV) and oxidized to manganese(VII).
#stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) > stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^() + stackrel(color(blue)(+4))("Mn")"O"_ (2(s))#
The reduction halfreaction looks like this
#stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) + 2"e"^() > stackrel(color(blue)(+4))("Mn")"O"_ (2(s))#
Here each atom of manganese takes in
To balance the atoms of oxygen, use the fact that this reaction takes place in an acidic medium and add water molecules to the side that needs oxygen and protons,
#4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) + 2"e"^() > stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))#
Notice that the halfreaction is balanced in terms of charge because you have
#4 xx (1+) + (2) + 2 xx (1) = 0 + 0#
The oxidation halfreaction looks like this
#stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) > stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^() + "e"^()#
This time, each atom of manganese loses
The atoms of oxygen are already balanced, so you don't need to use water molecules and protons. Once again, the halfreaction is balanced in terms of charge because you have
#(2) = (1) + (1)#
So, you know that the balanced halfreactions look like this
#{(4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) + 2"e"^() > stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))), (color(white)(aaaaaaaaaaaaaa)stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) > stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^() + "e"^()) :}#
In every redox reaction, the number of electrons lost in the oxidation halfreaction must be equal to the number of electrons gained in the reduction halfreaction, so multiply the oxidation halfreaction by
#{(4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) + 2"e"^() > stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))), (color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) > 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^() + 2"e"^()) :}#
Add the two halfreactions to find the balanced chemical equation that describes this disproportionation reaction.
#{(4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) + 2"e"^() > stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))), (color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2) > 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^() + 2"e"^()) :}#
#color(white)(a)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#
#4"H"_ ((aq))^(+) + ["MnO"_ (4(aq))^(2) + 2"MnO"_ (4(aq))^(2)] + color(red)(cancel(color(black)(2"e"^()))) > 2"MnO"_ (4(aq))^() + "MnO"_ (2(s)) + color(red)(cancel(color(black)(2"e"^()))) + 2"H"_ 2"O"_ ((l))#
You will end up with
#4"H"_ ((aq))^(+) + 3"MnO"_ (4(aq))^(2) > 2"MnO"_ (4(aq))^() + "MnO"_ (2(s)) + 2"H"_ 2"O"_ ((l))#
I get approximately
NOTE: The answer is kind of long!
We have a sphere of sterling silver with a radius of
The volume of a sphere is given by:
where
Plugging that into the equation, we get
So, the volume of this alloy in the shape of a sphere is
We know that
So,
The density of this alloy is
So, this alloy's mass is
Since
To find the amount of copper atoms in this alloy, we first need to find the number of moles of copper, and we can do that by dividing the total mass of copper atoms in here by copper's molar mass.
Copper
Therefore, there are
To convert between moles to atoms, we multiply the number of moles by Avogadro's number
So we have in total:
(i)
This tells you that the van der Waals equation of state picks up on the attractive interactions in acetylene that the ideal gas law assumes is not present.
(a) Ideal gas
The equation for the Ideal Gas Law is
#color(blue)(bar(ul(color(white)(a/a)pV = nRTcolor(white)(a/a))))" "#
Then,
#p = (nRT)/V#
The pressure predicted by the Ideal Gas Law is 4.70 atm.
(b) van der Waals Equation
The van der Waals equation is
#color(blue)(bar(ul(color(white)(a/a) (P + (n^2a)/V^2)(V  nb) = nRTcolor(white)(a/a))))" "#
#P + (n^2a)/V^2 = (nRT)/(V  nb)#
#P = (nRT)/(V  nb) (n^2a)/V^2#
For this problem,
The pressure predicted by the van der Waals equation is 4.58 atm, lower than for the ideal gas law.
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