Calculating pH

[phunkeyfantom]

Carbonic acid has Ka of 4.3 x 10^7. What is the pH when 1 mole of NaHCO3 is dissolved in 1 liter of water?

A. 3.2
B. 3.8
C. 10.2
D. 12.5

Completely lost as to how this problem was solved in the EK chemistry book since they used Kb to calculate the pH. According to them, the answer is C. I'm not sure where they got any of their numbers from and I'm wondering if this maybe an error because when I did the calculations, I got choice A as the pH. This is from just using the information given. Any help you guys can provide would be greatly appreciated. Thanks.

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[brianphu53]

Can you tell me what ek book said?

 

[phunkeyfantom]

Can you tell me what ek book said?

According to the EK answer key, it says that the correct answer choice is C. They solved the problem using Kb

set up the solution as such

Kb = [OH-][H2CO3] / [HCO3]

0.25x10^-7 = [x][x]/ [1-x]
2.5 x10^8 = x^2
1.5 x10^-4 = x

The pOH = btw 3 & 4. Subtract from 14, the pH = btw 10 & 11.

Can you see where my confusion stems from? I don't understand why they even bothered to give us the Ka value if they were going to be using a Kb value to solve the problem. The issue is that I have no idea where they got a Kb value of 0.25x10^7 from or the motivation behind using Kb.

 

[adrakdavra]

We expect a solution of NaHCO3 to be basic since it comes from NaOH (strong base) and H2CO3 (weak acid). so you can eleminate A and B, and Ka = X2 = 2x10^3.5 , so the PH is 3.5 for Ka, but because it is a base we need the POH which is 14-3.5 = 10.5. , so the first thing I would do is to try to know if it is an acid or base

 

[phunkeyfantom]

Thanks so much for your response. I literally spent the entirety of yesterday being haunted by how to correctly solve this problem.

 

[Gust]

According to the EK answer key, it says that the correct answer choice is C. They solved the problem using Kb

set up the solution as such

Kb = [OH-][H2CO3] / [HCO3]

0.25x10^-7 = [x][x]/ [1-x]
2.5 x10^8 = x^2
1.5 x10^-4 = x

The pOH = btw 3 & 4. Subtract from 14, the pH = btw 10 & 11.

Can you see where my confusion stems from? I don't understand why they even bothered to give us the Ka value if they were going to be using a Kb value to solve the problem. The issue is that I have no idea where they got a Kb value of 0.25x10^7 from or the motivation behind using Kb.

the Ka or Kb can be found by using the formula 1x10^-14=[Ka][Kb]...but to answer the question.
1)find the molarity which is 1mole/1liter= 1M of carbonic acid
2)set up an ice table for the reaction
H2CO3 H2O <-> H3O+ HCO3-
Initial concentration 1M n/a 0 0
Change concentration -x n/a +x +x
Equilbrium concentration 1M-x n/a x x
3)set up the equation and plug in values
Products/reactants=Ka

[H3O+]^1[HCO3-]^1/[H2CO3]^1=Ka
[x]^1[x]^1/[1-x]^1=4.3*10^-7 u can drop the -x from the denominator
which leaves u with the expression [x]^2/[1]=4.3*10^-7

solve for x...this x value equals 6.55*10^-4...looking back at your ice table u see that x is the concentration of H3O+ (H+) concentration...

plug in value into -log[H3O+]=pH and ur answer is 3.18.

BUT, the work ive shown u is for carbonic acid (H2CO3), the question is asking u about the conjugate base(HCO3-)...so the question is saying u have 1M of HCO3- not H2CO3. so u have to do the same thing ive shown above now using HCO3- but also using the Kb.

1)find the molarity which is 1mole/1liter= 1M of HCO3-
2)set up an ice table for the reaction
HCO3- H2O <-> OH- H2CO3
Initial concentration 1M n/a 0 0
Change concentration -x n/a +x +x
Equilbrium concentration 1M-x n/a x x
3)set up the equation and plug in values
Products/reactants=Kb

[OH-]^1[H2CO3]^1/[HCO3-]^1=Kb
[x]^1[x]^1/[1-x]^1=2.3*10^-8 u can drop the -x from the denominator
which leaves u with the expression [x]^2/[1]=2.3*10^-8

solve for x...this x value equals 1.52*10^-4..looking back at your ice table u see that x is the concentration of OH-concentration...

plug in value into -log[OH-]=pOH and ur answer is 3.81.

to get the pH, use pH+pOH=14 and ur pH will be 10.19

 

[adrakdavra]

Good Luck

 

[Tarasaurusrex]

Carbonic Acid is: H2CO3, that is our acid. Our conjugate base is NaHCO3. So now set up an equation that will relate the two, knowiing that this is an acid-base equation so an H is going to be added on one side, lost on the other. The side that can accept the proton is a base, the donating side is the acid. Also remember we are replacing a strong acid or base, in this instance, with a weak one.

H2CO3 + NaOH <----> OH- + NaHCO3-

Use the ICE table where x represents the OH- ion

(1 - x) x x

Since it's a hydroxide ion we are going to use Kb, so Kb = Kw/Ka, Kw = 1.0 x 10^-14/4.3 x 10^-7 = 2.326 x 10^-8

Kb = x^2/(1-x) = 2.326 x 10^-8

The x in the denominator is negligble, so solve for x by taking the square root

x = 1.525 x 10^-4
-log(x) = pOH = 3.817
14 - 3.817 = pH
pH = 10.18 or rounded, is 10.2

 

[circulus vitios]

According to the EK answer key, it says that the correct answer choice is C. They solved the problem using Kb

set up the solution as such

Kb = [OH-][H2CO3] / [HCO3]

0.25x10^-7 = [x][x]/ [1-x]
2.5 x10^8 = x^2
1.5 x10^-4 = x

The pOH = btw 3 & 4. Subtract from 14, the pH = btw 10 & 11.

Can you see where my confusion stems from? I don't understand why they even bothered to give us the Ka value if they were going to be using a Kb value to solve the problem. The issue is that I have no idea where they got a Kb value of 0.25x10^7 from or the motivation behind using Kb.

For any conjugate acid-base pair, Ka * Kb = Kw = 1E-14 aka the dissociation constant of water.

Kb = (1E-14)/(4.3E-7) ~= 2.33E-8.

I like to use the TBR shortcut for the Henderson Hasselbach equation:
pH = pKa/2 - log([HA])/2
pOH = pKb/2 - log([BOH])/2

pKb = -log(Kb) = 7.63
[BOH] = (1 mol HCO3)/(1 L water) = 1 mol

pOH = 7.63/2 - log(1 M)/2 = 3.82
pH = 14 - pOH = 10.18

 

[LoLCareerGoals]

Do we expect to get the same answer using Ka2 of the carbonic acid (if it was given in the problem as well)? We start with NaHCO3, why you guys are bringing up NaOH and H2CO3 as starting reactants is unclear to me when the problem does not even mention them.

NaHCO3 -> [Na+] + [HCO3-], molarity of [HCO3-] = 1M
[HCO3-] <-> [H+] + [CO3 2-], ka = 5.6E-11
[H+]^2 = 5.6E-11
ph = -log10[H+] or 2*ph = -log10(5.6E-11)
ph = 5.12


Why am I not getting the same answer as you guys?

 

[Tarasaurusrex]

LOLCareerGoals,

-log10(5.6E-11) = 10.25....so you got the same answer as everyone else, you must have just mistakenly hit something wrong on your calculator....?

The reason I bring up the NaOH and H2CO3 is because I am extremely visual and it helps me visualize the breakdown and assign an x-value to the (H3O+) or in this case the OH- ion, which is integral in solving this problem. I'm just very visual, kudos if you can do it without the visualizations!

 

[LoLCareerGoals]

Well no, in my case 2*ph = 10.25...

I am trying to understand how to treat ions of multiprotic acids. Why do we automatically assume it would act as a base is what I am not 100% clear on.
Is HCO3- simultaneously an acid and a base in water? Are there 2 processes happening at the same time?
1st: HCO3- <-> H+ + CO3^2-
2nd: HCO3- + H3O+ <-> H2CO3 + H2O
but only the effect of 2nd is more dominant in its effect on pH? How can you tell?