Acid Base Buffer Solutions Questions

[WellingtonSears]

Hi, excuse me if this isnt the right place. Its been a while since coming here and the layout has changed considerably. If its not, ill change it, but I have 2 questions regarding buffer solutions.

What is [H3O] in a solution of .075 M HNO2 and .030 M NaNO2

HNO2 + H2O --> H3O + NO2 Ka= 4.5 x 10^-5

and...

Calculate the PH of a buffer solution prepared from .155 mol phosphoric acid and .25 mol KH2PO4 in enough water to make 500 L of solution.

Am I correct in thinking that you would use the common ion effect to solve these problems? For instance, in the first problem, since the conjugate base is NO2 then you couldnt just plug into the Ka formula to find the answer as the conjugate base is NO2, but NaNo2 adds a common ion so Id use an ICE table with x values, right?

And the second problem would use the HH since it asks for PH, after conversions and all, id also make an ICE table with x values. Id also have to look up Ka, right? As its not provided, theres no way to really calculate it.

Sorry, its been a while since doing any of this, so I was wondering if I was on the right track here. Thank you in advance!

---

Members don't see this ad.

 

[NextStepTutor_4]

You're definitely on the right track.

For the first problem, you're exactly right: this is a common ion problem, and we'll need to use an ICE table and the Ka equation to solve.

1. First, set up the reaction
HNO2 + H2O --> H3O+ + NO2-

2. Next, set up an ICE table. We start with 0.075 M HNO2; we're always going to ignore water; no H3O+; and 0.03 M NO2-. Then we lose x amount of our reactants while gaining x amount of products.

..........HNO2 + H2O --> H3O+ + NO2-
I.........0.075...................0.........0.03
C........... - x..................+ x........+ x
E.......0.075 - x................x.......0.03 + x

3. Now we can plug our equilibrium values in the Ka expression, since this is a weak acid:

Ka = [H3O+][NO2-]/[HNO2] = (x)(0.03+x)/(0.075 - x)

4. It turns out that "x" is going to be so small compared to 0.03 or 0.075 M that it's negligible, and we can toss x out in 0.03 + x and 0.075 - x:

Ka = 4.5 x 10^-5 = (x)(0.03)/(0.075)

5. Now we can solve!

4.5 x 10^-5 = (x)(0.03)/(0.075)
3.375 x 10^-6 = 0.03 x
x = 1.125 x 10^-4 M

Remember: x tells us the molar concentration of H3O+ (which is essentially how H+ actually exists in water). So the answer is 1.125 x 10^-4 M. Does that make sense?

 

[NextStepTutor_4]

For your second question, we definitely are going to need a Ka value, since phosphoric acid (H3PO4) is a weak acid. The problem, however, is that it's triprotic, so it has 3 KA values, one for each proton. Which one should we use? Well, it turns out that the first proton dissociates much, much more readily than the other 2 protons, so we can treat phosphoric acid like a monoprotic acid and just use that first Ka1 value. We're going to solve like before:

1. Set up the equation:

H3PO4 + H2O --> H3O+ + H2PO4-

2. We're given number of moles and the volume of solution, but we need to convert these to molar concentration:

0.155 mol / 0.5 L = 0.31 M H3PO4
0.25 mol / 0.5 L = 0.5 M H2PO4-

3. Set up an ICE table:

.......H3PO4 + H2O --> H3O+ + H2PO4-
I........0.31.....................0..........0.5
C.........- x...................+ x..........+ x
E.....0.31 - x..................x.........0.5 + x

4. Plug into our Ka expression:

Ka1 = [H3O+][H2PO4-]/[H3PO4] = (x)(0.5+x)/(0.31-x)

5. Let's drop those x values since they are negligible for this weak acid:

Ka1 = 7.1 x 10^-3 = (x)(0.5)/(0.31)

6. Let's solve:

2.201 x 10^-3 = 0.5x
x = 4.402 x 10^-3 M H3O+

7. x tells us the molar concentration of H3O+, but we're asked to find the pH. We have to use the pH equation:

pH = -log[H3O+] = -log(4.402 x 10^-3) = ??

...but on Test Day, you won't be able to plug logarithms into your calculator! So let's estimate. 4.402 x 10^-3 is somewhere between 10^-2 and 10^-3. So -log(10^-2) is just 2 and -log(10^-3) is just 3, so our answer must be somewhere between 2 and 3, in other words a really acidic pH. In fact, the actual pH is 2.36.

 

[WellingtonSears]

Awesome! That really helps, thank you so, so much.

 

[NextStepTutor_4]

Awesome! That really helps, thank you so, so much.

You're welcome!

 

[WellingtonSears]

Could you answer just one more question? Which will increase % of HF that's converted to F ion in water?
a.) Adding NaOh
b.) Adding HCl
c.) Adding NaF

Would it be A due to Le Chatliers principle? Adding HCl and NaF would increase both the H and F concentrations, which would cause an increase in HF to compensate, right? But adding NaOH would cause it to dissociate more?

 

[NextStepTutor_4]

Could you answer just one more question? Which will increase % of HF that's converted to F ion in water?
a.) Adding NaOh
b.) Adding HCl
c.) Adding NaF

Would it be A due to Le Chatliers principle? Adding HCl and NaF would increase both the H and F concentrations, which would cause an increase in HF to compensate, right? But adding NaOH would cause it to dissociate more?

Exactly, this is a Le Châtelier's problem, so I like writing out the equation so I can visualize what's happening at each step.

HF + H2O --> H3O+ + F-

Let's start with an easier one (B). If we add acid (HCl), we're essentially adding more protons (H3O+), so in which direction with the reaction shift? Towards the left, so less dissociation will occur.

Now let's try the opposite scenario (A). If we add base, well we don't exactly see OH- ions in this equation. But what do you think OH- will react with? It's going to neutralize acid, so OH- will combine with H3O+ to form H2O. If we consume H3O+, a product, the reaction will shift to the right, and more dissociation will occur.

Last but not least (C), if we add a common ion (F- in NaF), we're essentially adding more product. To return to equilibrium, the reaction will shift to the left, and less dissociation will occur.

So you're exactly right!

 

[WellingtonSears]

Perfect, youve been very helpful - thank you