PH - really basic, I know, but I'm tired

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Discussion Overview

The discussion revolves around calculating the pH of a solution containing sodium carbonate and sodium bicarbonate. Participants explore the implications of these compounds in a buffer solution context, addressing the necessary calculations and chemical principles involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the calculation of the concentration of sodium bicarbonate and expresses uncertainty about the resulting pH value, suggesting a possible error in their approach.
  • Another participant emphasizes that the final volume does not affect the buffer solution calculations and suggests using the Henderson-Hasselbalch equation with the pKa value for carbonic acid.
  • There is a reiteration of the need to calculate the number of moles of each compound and apply the stoichiometric ratios to find the concentrations.
  • A later reply provides a detailed calculation using the Henderson-Hasselbalch equation, presenting a specific pH value while acknowledging assumptions made in the calculations.

Areas of Agreement / Disagreement

Participants express various viewpoints on the calculations and methods to determine pH, indicating that there is no consensus on the best approach or the correctness of the calculations presented.

Contextual Notes

Some participants note the importance of considering the weak acid equilibrium constant and the stoichiometric ratios, while others express confusion about the calculations and the implications of the final volume in the context of buffer solutions.

Who May Find This Useful

This discussion may be useful for individuals interested in buffer solutions, pH calculations, and the application of the Henderson-Hasselbalch equation in chemistry.

CasanovaFrankenstein
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pH -- really basic, I know, but I'm tired

Hi Chem masters out there,

Anyone want to help me with this one?

What is the pH of a solution with 10 g of sodium carbonate and 10 g of sodium bicaronate dissolved in enough water to make 0.25 mL of solution?
 
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Sodium Carbonate = [tex]Na_2CO_3[/tex]
Sodium Bicarbonate = [tex]NaHCO_3[/tex]

0.25 mL = 0.0025L

You can ignore the Na+ ion because it's neutral, and the sodium bicarbonate releases the H proton into solution whih results in the hydronium ion being formed, therefore the concentration of the sodium bicarbonate will be equal to the concentration of the H+ ion.

Concentration of sodium bicarbonate:
10g
molar mass = 84.01 g/mol
to get moles, take grams and divide it by molar mass:
moles of sodium bicarbonate:
[tex]1.2 times 10^{-1}[/tex]

divide the moles over litres of solution to get concentration:

47.6 M? okay that can't be right =/

Then you're supposed to -log(x) that to get the pH.
In case of a base, get the pOH then 14 - pOH to get pH.
I think I did something wrong. Maybe my chemical formula is wrong? I think it might form OH- actually =(

I'm sorry!
 
Last edited:
Borek said:
This is the buffer solution, so the final volume doesn't count. Calculate number of moles of both compounds and put them into HH equation.

pKa2 for carbonic acid is 10.25.


Borek
--
*gasps*
I'm stupid =(
 
just in case "HH" is the henderson hasselbach equation.

Note that 10g represents the whole ionic compound, you'll need to first find the molar mass of each compound, then convert 10g to moles. Then you'll need to use the stoichiometric ratio (which may be 1:1, you should figure it out), to find the moles of each anionic component and divide each value by the volume. Remember to use the correct units.
 
Thanks Guys! That was fast and easy
 
Artermis said:
Sodium Carbonate = [tex]Na_2CO_3[/tex]
Sodium Bicarbonate = [tex]NaHCO_3[/tex]

0.25 mL = 0.0025L

You can ignore the Na+ ion because it's neutral, and the sodium bicarbonate releases the H proton into solution whih results in the hydronium ion being formed, therefore the concentration of the sodium bicarbonate will be equal to the concentration of the H+ ion.

Concentration of sodium bicarbonate:
10g
molar mass = 84.01 g/mol
to get moles, take grams and divide it by molar mass:
moles of sodium bicarbonate:
[tex]1.2 times 10^{-1}[/tex]

divide the moles over litres of solution to get concentration:

47.6 M? okay that can't be right =/

Then you're supposed to -log(x) that to get the pH.
In case of a base, get the pOH then 14 - pOH to get pH.
I think I did something wrong. Maybe my chemical formula is wrong? I think it might form OH- actually =(

I'm sorry!

I like the way you solved it lol, you have to incorporate the weakacid equilibrium constant..
HCO3- => H+ + CO3^2- with Ka2 of 1.5 x 10^-11
you have (10.g)/(84.0g/mol NaHCO3) = .12mol HCO3-
and have (10.g)/(106.0g/mol Na2CO3) = .094mol CO3^2-
.12mol HCO3- / .00025L = 480M
.094mol CO3^2- / .00025L = 376M
Using Henderson-Hasselbalch Equation.. pH = pKa + log( [CO3^2-]/[HCO3-] )
you get.. pH = -log(1.5E-11) +log(376M/480M) assuming 376+x/480-x ~= 376/480.
so you will get: pH = 10.82 - .106 = 10.71 thus pH=10.71
if you don't understand something i did, tell me :p
 

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