Calculating the pH of a Buffer Solution

  • Thread starter DapperDan
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Problem Statement
(1) You are supplied with propanoic acid, CH3CH2CO2H, which has a pKa of 4.87 and density of 0.992 g/mL. If you take 8.62 mL of propanoic acid and make an aqueous solution with a total volume of 1450.0mL, what is the molarity of the solution?

(2) What is the pH of the propanoic acid solution?

(3) If 9.6 grams of sodium propanoate is added to 1.00L of the propanoic acid solution made in (1), what is the new pH for the buffer solution?

(4) If 0.01M of HCl(g) is bubbled into the buffer solution in (3), what is the change in pH assuming that there is no volume change?

(5) If 0.40g of NaOH(s) is dissolved in the buffer solution in (3), what is the change in pH assuming that there is no volume change?
Relevant Equations
Ka=[H+][A-]/[HA]
pH = -log[H+]
Others?
Hi everyone. I'm doing the multi-step problem above, and I've found myself stuck at part 3.

For step 1, I determined that the equation that we're dealing with is:
$$CH_3CH_2CO_2H + H_2O ⇔ CH_3CH_2CO_2^- + H_3O^+$$

I've also determined that the molar mass of propanoic acid is about 74.08g/mol, and we therefore have 0.11543 mol of propanoic acid. So we have $$\frac{0.11543 mol}{1.45L} = 0.07961 M$$

For part (2) I first found Ka. We are given pKa as 4.87. pKa = -log(Ka) so $$-log(K_a)=4.87$$
$$log(K_a)=-4.87$$
$$K_a=10^{-4.87}$$
$$K_a=1.349*10^{-5}$$

Based on the equation I determined in the previous step, we know that $$K_a=\frac{[CH_3CH_2CO_2^-][H_3O^+]}{[CH_3CH_2CO_2H]}$$
Using an ICE table I determined that $$1.349*10^{-5}=\frac{[x][x]}{[0.07961-x]}$$.
When solved for x, this produces x=0.00103, which corresponds to the concentrations of H30+ and CH3CH2CO2-. As pH is equal to -log[H+], we get pH=-log(0.00103), which gives pH=2.987.

For part 3, however, I am completely stumped. I cannot begin to even understand what our chemical equation would look like for the reaction involved sodium propanoate with the propanoic acid solution. I found the forumla for sodium propanoate, Na(C2H5COO), so I have $$CH_3CH_2CO_2^-+H_3O^+ +Na(C_2H_5COO) ⇔ ???$$

But I don't know how to complete this equation. How do these compounds interact? How can I write the equation for this interaction so I can start on this part of the problem? Please help, many thanks in advance for any assistance that can be provided.
 

Borek

Mentor
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Hard way: salt is dissociated, you can use ICE table with known initial concentration of propanoate anion.

Easy way: use Henderson-Hasselbalch equation. Technically it is just a rearrangement of the dissociation constant, but it is very useful for buffer problems.

 
Thank you, I'm always down to use easy solutions!

For point of reference though, what would the dissociation of the salt look like? How would that be written in this case?
 
Also I have made more progress on the problem I think, if anyone is willing to take a look at my work:

In part (3) we are given 9.6 grams of Na(C2H5COO) which amounts to 0.0999349382 or roughly 0.1 moles. In one liter of solution we get 0.1M. Then using the Henderson-Hasselbalch equation:
$$pH=pK_a + log(\frac{[A-]}{[HA]})$$
$$pH=4.87+log(\frac{[0.1]}{[0.07961]})$$
$$pH=4.87+0.099032376$$
$$pH=4.97$$

For part (4) we are given 0.01M HCl which would give us 0.01M Cl-, our conjugate base. So then:
$$pH = pK_a+log(\frac{[A-]}{[HA]})$$
$$pH=4.87+log(\frac{[0.01]}{[0.07961]})$$
$$pH=4.87-0.9009676239$$
$$pH=3.97$$

Does this look correct?
 

Borek

Mentor
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For part (4) we are given 0.01M HCl which would give us 0.01M Cl-, our conjugate base.
Doesn't sound correct to me - what is the acid, what is the conjugate base in this buffer?

What is happening is that H+ from HCl protonates the propanoate anion, removing the conjugate base. You can assume protonation went to completion (it is not exact, but close enough).

Miraculously, your result looks OK.
 

epenguin

Homework Helper
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509
For part 4 Protonation has just decreased [A-] by 0.01 relative to part 3, and increased [HA] by the same amount. So I think you have to re-calculate numerator and denominator in the HH equation.

Part 5 should not be difficult, I think it is practically just returning to a previous situation.
 
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