Calculating the pH of a Buffer Solution

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Discussion Overview

The discussion revolves around calculating the pH of a buffer solution involving propanoic acid and sodium propanoate. Participants explore various methods for determining pH, including the use of the Henderson-Hasselbalch equation and ICE tables, while addressing specific steps in a multi-part problem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the dissociation of propanoic acid and calculates its concentration and Ka value, leading to an initial pH calculation.
  • Another participant suggests using the Henderson-Hasselbalch equation as a simpler method for calculating pH in buffer problems.
  • A participant provides calculations for the pH after adding sodium propanoate, using the Henderson-Hasselbalch equation with given concentrations.
  • Concerns are raised about the identification of the acid and conjugate base when hydrochloric acid is introduced, questioning the accuracy of the previous calculations.
  • Participants discuss the effects of protonation on the concentrations of the acid and conjugate base in the buffer solution.
  • One participant reflects on a related experience with buffer solutions, noting practical challenges encountered in a laboratory setting.
  • Another participant critiques the reasoning behind the pH change calculations, emphasizing the need for reasonable assumptions in buffer behavior.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions made regarding the buffer system, particularly concerning the effects of adding HCl and the resulting pH changes. There is no consensus on the correctness of the calculations or the interpretation of the buffer dynamics.

Contextual Notes

Participants highlight the importance of understanding the interactions between the acid and its conjugate base, as well as the implications of adding strong acids to the buffer solution. There are unresolved aspects regarding the exact changes in concentrations and the resulting pH calculations.

Who May Find This Useful

This discussion may be useful for students and practitioners in chemistry, particularly those studying buffer solutions and pH calculations in laboratory settings.

DapperDan
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Homework 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 formula 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.
 
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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.

http://www.chembuddy.com/?left=buffers&right=toc
 
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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?
 
DapperDan said:
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.
 
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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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Slightly off-topic: The example bears a horrid resemblance to the bizarre buffer we had to use to adjust a smorgasbord HPLC mobile phase for a {redacted} pharmaceutical cogener / impurity analysis.

IIRC, despite our 'good practice', it progressively contaminated pH electrodes, took up atmospheric CO2 fast enough to be scary, and grew 'bugs' quicker than an exposed agar plate...
YMMV.
 
For part 4 I get the ΔpH to be close to -0.1 following the indications I gave of how to calculate it.

More important than an answer being right or wrong, is to realize whether it is reasonable or not according to rough reasoning.

The solution has a pH near its pKa. To bring the pH down by one unit, as claimed, you would have to protonate of the order of 90% of the proprionate ions present which are 0.1 M, and there has been added only a 10th of their amount of acid.

More than a week has passed since the last post so sadly it looks the OP will have to go into my black book, see my sig.
 
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