calculating pH

[Oh_Gee]

from GS-3


"Cyanide ions (CN-) could also have been used to precipitate the cations. If after all the cations have been precipitated, the concentration of CN- is 0.02 M, calculate the pH of the solution given that:
CN- + H2O --> HCN + OH-
Kb = 1.39 x 10-5

a. 4.9
b. 5.4
c. 7.7
d. 10. 7

Using Kb = [HCN][OH-]/[CN-]

Assuming that [HCN] = [OH-] approximately

Kb = [OH-]2/[CN-]

[OH-]2 = Kb x [CN-] = 1.39 x 10-5 x 0.02 = 1.39 x 10-5 x 2 x 10-2 = 2.78 x 10-7

[OH-] = square root (2.78 x 10-7) = square root (27.8 x 10-8)

[OH-] = (square root 27.8) x 10-4 = (square root 25) x 10-4 approximately = 5 x 10-4

Using pOH = -log[OH-]

pOH = -log(5 x 10-4) = -log(5) [-log(10-4)] = -log(5) 4

{To see how a log can be estimated within reasonable error, see the end of CHM 6.6.1}

pOH = 3.5 approximately

Using pH + pOH = 14

pH = 14 - pOH = 14 - 3.5 approximately = 10.5"


what I did was solve for ka via ka x kb = 10^-14

and then take negative log of ka to get pH

but then I realized that you can't directly get pH from pKa can you?

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[justadream]

@Oh_Gee

What's the answer?

 

[Teleologist]

Assuming that [HCN] = [OH-] approximately

Why does every single student of chemistry except me always say assume that the above is true? It is true to the number of sig figs we're worried about in this problem.

but then I realized that you can't directly get pH from pKa can you?

Of course you can calculate pH from pKa.

 

[type12]

The only time the pka and pH are one in the same is at the half equivalence point. The pka is NOT the pH, but a constant value you can use to get the concentration of H:

k_a = [H+][A-]/[HA]

When you take the log of this equation, and move values around:

log(k_a) = log[H+] + log [A-]/[HA]
-log[H+] = -log(k_a) + log[A-]/[HA]
pH = pka + log[A-]/[HA] (how I wish SDN had LaTeX)

You see that when log[A-]/[HA] is zero, pH = pka, and that's only when concentrations of the acid and the conjugate base are equal (aka the half equivalence point).

I would try to avoid the Henderson-Hasselhoff version of the equation until you understand the acid disassociation constant. So you can see that when A- and HA cancel each other out:

k_a = [H+] * 1 / 1

the k_a is equal to the concentration of H ions, which means pH and pka are equal. So, you could do your approach just fine with ka, as you can figure out the values of ka, CN-, and HCN, and thereby solve for H+, getting you the pH.

 

[Oh_Gee]

@Oh_Gee

What's the answer?

GS says it is D

"Using Kb = [HCN][OH-]/[CN-]

Assuming that [HCN] = [OH-] approximately

Kb = [OH-]2/[CN-]

[OH-]2 = Kb x [CN-] = 1.39 x 10-5 x 0.02 = 1.39 x 10-5 x 2 x 10-2 = 2.78 x 10-7

[OH-] = square root (2.78 x 10-7) = square root (27.8 x 10-8)

[OH-] = (square root 27.8) x 10-4 = (square root 25) x 10-4 approximately = 5 x 10-4

Using pOH = -log[OH-]

pOH = -log(5 x 10-4) = -log(5) [-log(10-4)] = -log(5) 4

{To see how a log can be estimated within reasonable error, see the end of CHM 6.6.1}

pOH = 3.5 approximately

Using pH + pOH = 14

pH = 14 - pOH = 14 - 3.5 approximately = 10.5"

 

[justadream]

@Oh_Gee

I'm not really sure how to use pKa/Ka to solve this but if I were doing this problem myself, I'd just use the given Kb (the GS solution).

Here's my work (I think it's basically the same as the GS solution).

Just set up the ICE chart and you will get
CN- + H2O --> HCN + OH-
.02 <=> +x +x

Kb = 1.39 x 10-5 = [HCN][OH-] / [CN-]

Kb = 1.39 x 10-5 = x^2
Solve for x and get x = 5.27x10^-4
x = [OH-] =5.27x10^-4

To find pOH, do -log(x) = 3.3

pH = 14-3.3 = 11.3

 

[Oh_Gee]

Why does every single student of chemistry except me always say assume that the above is true? It is true to the number of sig figs we're worried about in this problem.


Of course you can calculate pH from pKa.

1. The answer explanation was quoted verbatim from the GS explanation. I don't agree with a lot of their explanations either lol

2. ph= -log Ka

right?
could you quick fire all the ways to convert between ph, ka, pka? I can't seem to keep them all in my mind

 

[Teleologist]

2. ph= -log Ka

right?
could you quick fire all the ways to convert between ph, ka, pka? I can't seem to keep them all in my mind

Try writing out what Ka equals for a generic acid, say, HA, and then take the negative log of both sides, and rearrange until you get pH = pKa + log[base]/[acid], and then think about when pH might equal pKa.

 

[Oh_Gee]

Try writing out what Ka equals for a generic acid, say, HA, and then take the negative log of both sides, and rearrange until you get pH = pKa + log[base]/[acid], and then think about when pH might equal pKa.

is my equation valid though?

 

[Teleologist]

is my equation valid though?

Only sometimes

 

[Oh_Gee]

Only sometimes

when???

 

[Teleologist]

when???

Try doing what I suggested

 

[Chrisz]

Try writing out what Ka equals for a generic acid, say, HA, and then take the negative log of both sides, and rearrange until you get pH = pKa + log[base]/[acid], and then think about when pH might equal pKa.

@Oh_Gee, pH=pKa+log(base/acid) does not apply in this type of questions, it only applies to buffer type questions when you know the ratio base/acid. In this question, you do not know what is ratio at the first place. However, you can use another equation to calculate the pOH from pKb and then calculate pH from pOH.
pOH= 0.5pKb-0.5 log(base)

The logic behind this is Kb=[HBase][OH]/[Base] since you only have Base initially and you will generate equal amount HBase and OH as the reaction proceeds, so
Kb=[OH]^2/[base]
pKb=2pOH-p[base]
pKb+p[Base]=2pOH
0.5(pKb-log[base])=pOH

Lets apply this equation to see if we get the right answer.
Kb=1.39x10^-5
pKb=5-log(1.39)
aprox=5-log(3/2)
=5-(0.47-0.3)
=5-0.17=4.83
pOH=0.5(4.83-log(0.02))
=0.5(4.83-(-2+0.3))
=0.5(4.83+1.7)
=3.25

pH=14-pOH=10.75

see this is exactly answer D
Edit:
earlier i used another notation for base, the system thought i wanted to bold the text. now it is fixed

 

[Chrisz]

The logic behind this is Kb=[HBase][OH]/[Base] since you only have Base initially and you will generate equal amount HBase and OH as the reaction proceeds, so
Kb=[OH]^2/[base]
pKb=2pOH-p[base]
pKb+p[Base]=2pOH
0.5(pKb-log[base])=pOH

Lets apply this equation to see if we get the right answer.
Kb=1.39x10^-5
pKb=5-log(1.39)
aprox=5-log(3/2)
=5-(0.47-0.3)
=5-0.17=4.83
pOH=0.5(4.83-log(0.02))
=0.5(4.83-(-2+0.3))
=0.5(4.83+1.7)
=3.25

pH=14-pOH=10.75

 

[Chrisz]

I don know why, it did not show the whole thing again. One more try
Kb=[HBase][OH]/[Base]
Kb=[OH]^2/[base]
pKb=2pOH-p[base]
(pKb+p[base])/2=pOH
0.5(pKb-log[base])=pOH

application
pKb=-log(1.39x10^-5)=4.83
pOH=0.5(4.83-log(0.02))=3.25
pH=14-3.25=10.75

This is answer D.

Edit: I earlier used another notation instead of [base]. the system thought i wanted to bold everything and omitted the notation. I will update the two ealier posts for better understanding

 

[Oh_Gee]

I don know why, it did not show the whole thing again. One more try
Kb=[HBase][OH]/[Base]
Kb=[OH]^2/[base]
pKb=2pOH-p[base]
(pKb+p[base])/2=pOH
0.5(pKb-log[base])=pOH

application
pKb=-log(1.39x10^-5)=4.83
pOH=0.5(4.83-log(0.02))=3.25
pH=14-3.25=10.75

This is answer D.

Edit: I earlier used another notation instead of [base]. the system thought i wanted to bold everything and omitted the notation. I will update the two ealier posts for better understanding

another equation to remember? 🙁
looks like it works well though

 

[Oh_Gee]

@Oh_Gee

I'm not really sure how to use pKa/Ka to solve this but if I were doing this problem myself, I'd just use the given Kb (the GS solution).

Here's my work (I think it's basically the same as the GS solution).

Just set up the ICE chart and you will get
CN- + H2O --> HCN + OH-
.02 <=> +x +x

Kb = 1.39 x 10-5 = [HCN][OH-] / [CN-]

Kb = 1.39 x 10-5 = x^2
Solve for x and get x = 5.27x10^-4
x = [OH-] =5.27x10^-4

To find pOH, do -log(x) = 3.3

pH = 14-3.3 = 11.3

when setting up ICE tables, how do you know when you can disregard the (-x) from the reactants side?

 

[justadream]

@Oh_Gee

I think there's a 5% rule (don't ask me how to calculate this - last time I did that was 5 years ago lol) but TBR has a chart where you can look at the Ka/Kb values and decide whether you can ignore it.

I think generally for MCAT purposes you can ignore it. Hopefully the answer choices will be far enough away that it doesn't really matter.

 

[Oh_Gee]

i remember that chart. it gave me nightmares but i should try thinking about why it works instead of trying to remember it...

 

[justadream]

@Oh_Gee

Correct me if I'm wrong but the basis is that the weak acid/base does not dissociate much. Thus, if you initially start with 1M, you remain at around 1M.

For MCAT purposes, I'd just ignore it.

 

[techfan]

Why does every single student of chemistry except me always say assume that the above is true? It is true to the number of sig figs we're worried about in this problem.

Because an = means EXACTLY equal, not approximately. Just because you maintain the correct amount of significant digits doesn't mean that something is equal; it's just a close approximation.

 

[techfan]

@Oh_Gee

I think there's a 5% rule (don't ask me how to calculate this - last time I did that was 5 years ago lol) but TBR has a chart where you can look at the Ka/Kb values and decide whether you can ignore it.

I think generally for MCAT purposes you can ignore it. Hopefully the answer choices will be far enough away that it doesn't really matter.

I think the chart you're referring to is if Keq is smaller than 10^-3 (and you start with all reactants) then x can be ignored because the reaction isn't moving very fast in the forward direction. Conversely if Keq is larger than 10^3 and you start with all products then x can be ignored because the reaction is moving very much in the reverse direction. Also if K_eq=1 then you can ignore x because the reaction is moving just as quickly in either direction.

 

[Teleologist]

Because an = means EXACTLY equal, not approximately. Just because you maintain the correct amount of significant digits doesn't mean that something is equal; it's just a close approximation.

Okay, if that's what you want to believe regarding significant figures, fine.

I think the chart you're referring to is if Keq is smaller than 10^-3 (and you start with all reactants) then x can be ignored because the reaction isn't moving very fast in the forward direction. Conversely if Keq is larger than 10^3 and you start with all products then x can be ignored because the reaction is moving very much in the reverse direction. Also if K_eq=1 then you can ignore x because the reaction is moving just as quickly in either direction.

Keq has nothing to do with how "fast" the reaction is moving in the forward direction. Acid-base reactions (proton-transfers) are generally VERY fast. Yet many acid-base reaction are small extent (many Ka values are < 1.0*10^-3). Let's not conflate kinetics with thermodynamics here.

Also if Keq = 1 that doesn't mean that there is no net reaction. That only applies for symmetric reactions, i.e. A + B -> C + D. Not for most acid-base reactions with water, since almost all are asymmetric.

 

[Chrisz]

@Teleologist, @techfan, I dont know why you two would argue on such trivia. To make use of an approximation, we have to make assumptions based on facts to make calculation easier. Of course, you dont wanna use -b+-sqr(b^2-4ac)/2a, and then write out the infinite number of digits without rounding to a smaller number of digits for a 100% accuracy. Since for weak base with high pKb and relatively high initial concentration, it does not react to a significant extent, we just make the assumption in our calculation the initial con of base constant. And water autoionization is ignorable and the common ion effect will further drive water autoionization to the left, so we can legitimately make another assumption that OH- is purely produced by the weak base in our calculation. Teleologist, it is a bit arrogant, condescending, mostly ignorant, to make your earlier statement,

Why does every single student of chemistry except me always say assume that the above is true? It is true to the number of sig figs we're worried about in this problem

When we assume something is true, it is not the same thing as we admit that something is true. We make assumptions based on a reasonable grounding, so as to make calculation easy and accurate to an acceptable extent.

 

[Teleologist]

@Teleologist, @techfan, I dont know why you two would argue on such trivia.

If you look at the content outline:

https://www.aamc.org/students/download/85562/data/ps_topics.pdf

Thermodynamics and kinetics are listed as separate topics. This suggests to me that the MCAT doesn't think that thermodynamics = kinetics.

Also the MCAT lists as one of its topics "dependence of reaction rate on temperature." I can't find anything talking about "dependence of reaction rate on K" though.

There is a difference between "it's not on the MCAT and we don't have to know it" (even though the distinction between thermo and kinetics should be an elementary topic and is on the MCAT) and "it's not on the MCAT so let's spread falsehoods and refuse to be corrected."

@TeleologistOf course, you dont wanna use -b+-sqr(b^2-4ac)/2a, and then write out the infinite number of digits without rounding to a smaller number of digits for a 100% accuracy. Since for weak base with high pKb and relatively high initial concentration, it does not react to a significant extent, we just make the assumption in our calculation the initial con of base constant. And water autoionization is ignorable and the common ion effect will further drive water autoionization to the left, so we can legitimately make another assumption that OH- is purely produced by the weak base in our calculation. Teleologist, it is a bit arrogant, condescending, mostly ignorant, to make your earlier statement,

No. We cannot make that "assumption" ([conjugate base] = [HO-]) when the weak base is dilute. The "assumption" has more to do with dilution factor rather than strength.

And water autoionization is ignorable and the common ion effect will further drive water autoionization to the left, so we can legitimately make another assumption that OH- is purely produced by the weak base in our calculation. Teleologist, it is a bit arrogant, condescending, mostly ignorant, to make your earlier statement,.

If you know all this, why are we still making assumptions? If you break down what you said, you're saying that the auto-ionization of water isn't affecting [HO-] in the solution. So we now have enough information to make a statement not an assumption, since there are only two sources of [HO-] in solution: the solute and water's auto-ionization. We ruled out the latter as negligible so now we don't have anything to assume ... the stoichiometry of the acid/base reaction is 1:1.

HA + H2O <-> H3O(+) + A(-)

 

[Chrisz]

@Teleologist, go back to the original poster's post. The original poster never stated the assumption that conjugate base=OH-. The original poster's assumption was conjugate acid =OH-. This assumption is valid for the original poster's question anyway.

why are we making the assumption not a statement of fact that water autoionization does have no effect. water auto indeed have an effect, but very small, negligible, that is why we make an assumption for calculation purpose. In fact, water dissociation takes place. In our calculation, it is inherently assumed that water does not affect the acid base reaction. This is the difference between assumption and statement of fact. If these assumption were not made, one would expect to see water auto to be incorporated in the original poster's equation. An assumption, whether implicitly or explicitly, has a different meaning from a statement of fact.

Since you was so worried about significant figures, that is why i brought up autoionization as an example.

Ok, lets stop here. I apologize if you was offended

 

[techfan]

Okay, if that's what you want to believe regarding significant figures, fine.



Keq has nothing to do with how "fast" the reaction is moving in the forward direction. Acid-base reactions (proton-transfers) are generally VERY fast. Yet many acid-base reaction are small extent (many Ka values are < 1.0*10^-3). Let's not conflate kinetics with thermodynamics here.

Also if Keq = 1 that doesn't mean that there is no net reaction. That only applies for symmetric reactions, i.e. A + B -> C + D. Not for most acid-base reactions with water, since almost all are asymmetric.

Sorry, I haven't taken a chemistry class (unless you count engineering thermodynamics) in years. Would you mind explaining (or posting a good link) on why the Keq isn't how fast the reaction is going? I'm using TBR and it made it seem like it was the reaction rate (which I equate to speed).

I don't think people here post on message boards very often, Teleologist's Dr. House "it doesn't matter how I express myself, as long as I'm right" attitude is the norm.

As for the Keq approximations I posted, I was simply trying to reinterpret a table (attached). Maybe I'm getting the concept wrong, and if so please explain (or point me to a source that can).

 

[Chrisz]

@techfan,
Thermodynamic only concerns where the energetic stability of products and reactants and if a reaction is favorable, while kinetics is concerned how fast a reaction proceeds.

Equilibrium constant tells you where an equilibrium position lies, but it does not tell you ho fast an equilibrium position will be reached.
Let me give you a an example of a one step concerted reaction
aA+bB-->cC+dD

Keq=(C^c*D^d)/(A^a*B^b)
at a specific temperature, there is a specific equilibrium constant, so given an initial amount of reactants, you can calculate the concentration of each reactant and product such that the equilibrium constant is satisfied when equilibrium position is reached.

For kinetics, here i am going to use some calculus notation (will later derive rate laws using calculus later. Pretty sure you can understand it since you was doing engineering back then)

The the formation of product should be at the same pace as the consumption of products
Therefore dB/dt=(b/a)dA/dt dC/dt=-(c/a)dA/dt dD/dt=-(d/a)dA/dt (negative signs included on the product side because formation is the opposite of consumption)

Rate Law:
For a generalized reaction aA-->bB

Forward Rate=k1A^a
Reverse Rate=k2B^b

k is called rate constant, k=Ae^(-Ea/RT)
e^(-Ea/RT) is the probability of a collision having energy equal to or greater than Ea, activation energy (in stats, it is called survival distribution)
A=pz is called arrheinius factor where p is the probability of a collision with the right orientation and z is the frequency of collision(# of collision per sec)

So the overall interpretation of k is the number of collision with the right collision orientation and collision energy equal to or greater than Ea



For a one step concerted reaction, equilibrium constant can be related to Kinetics

At equilibrium, the forward rate equals to reverse rate

Forw=Rev
k1A^a=k2B^b
k1/k2=B^b/A^a=Keq (this is the equilibrium constant expression)
e^(-Ea1/RT)/e^(-Ea2/RT)=Keq
e^{-(Ea1-Ea2)/RT}=Keq
See the equilibrium constant depends on the difference between the 2 activation energies and temperature.
So for a specific reaction, there is a specific Keq at a specific temp

Lets see how can we relate concentration with time (I will only derive the first order, for zero order and third order, the derivation is done in the same manner but different results

Rate Laws
(zero order reaction) dA/dt=-k
(first order) dA/dt=-kA
(second Order) dA/dt=-kA^2

Now lets derive the first order

dA/dt=-kA
dA/A=-kdt
lnA=-kt+c
at t=0, c=ln(A0)
lnA=-kt+ln(A0) This relates concentration with time

Now lets derive the half-life for the reaction
ln(A/A0)=-kt

at half life, we only have 1/2 of A0 left

ln(0.5Ao/A0)=-kt
ln(0.5)/-k=t
t=ln(2)/k

Lets derive another thing, which is interesting

ln(A/A0)=-kt
A/A0=e^-kt
This is another probability survival function with respect to time (this is the percentage of remaining reactants at time t)
Hope this helps