# Quick chemistry question

edited November 2010
I am doing pH and buffers right now, and in all my notes it says that to calculate pH to use molarity with Ka or Kb however I've just been using moles and I am getting the same answers. I have been doing this so that if it's a problem with the addition of a strong acid or base to a buffer I won't be confused as both my ICE tables will be in moles and not one in moles and one in molarity. Is there any reason to discontinue doing this?

[SIGPIC][/SIGPIC]
Otar on

## Posts

• edited November 2010
There are a few different things you might be confusing.

First, two quick definitions:
Molarity is the number of moles of the substance in one liter. This is a concentration.

[strike]Molality (note there are 2 "l"s). Is how much a weak acid or base dissociates as a 1 molar solution. This is a derived quantity, but it is also a concentration (i.e. the concentration of additional protons added to the solution by 1 mole of weak acid in 1 liter of water.) Depending on your grade level, you might not be covering this. Additionally, molality can change with concentration. If this isn't something you are going over, just ignore this.[/strike] Edit: This section is entirely incorrect, sorry!

What is important is that by using just moles, you are removing the volume from the equation. This may prove a problem if you have unequal volumes of multiple reagents. As with all stoichiometry problems, just keep track of your units and make sure the answer is in the correct units.

3clipse: The key to any successful marriage is a good mid-game transition.
• edited November 2010
There are a few different things you might be confusing.

First, two quick definitions:
Molarity is the number of moles of the substance in one liter. This is a concentration.

Molality (note there are 2 "l"s). Is how much a weak acid or base dissociates as a 1 molar solution. This is a derived quantity, but it is also a concentration (i.e. the concentration of additional protons added to the solution by 1 mole of weak acid in 1 liter of water.) Depending on your grade level, you might not be covering this. Additionally, molality can change with concentration. If this isn't something you are going over, just ignore this.

What is important is that by using just moles, you are removing the volume from the equation. This may prove a problem if you have unequal volumes of multiple reagents. As with all stoichiometry problems, just keep track of your units and make sure the answer is in the correct units.

We were taught to use molarity in our ICE tables for calculating the equilibrium concentration of hydronium or hydroxide and from there use that to get pOH or pH. But I have just been converting the volume of 0.010mol*L^-1HF or whatever to the number of moles found in said concentration and using moles in my ICE table. If there is a volume change I calculate the new concentration and get moles from that.

I can't seem to find anything wrong with it thus far I just wanted to ensure that this wasn't a no no for a reason that was not covered in lecture.

Otar on
[SIGPIC][/SIGPIC]
• edited November 2010
Molarity is moles per litre of solution(volume) while molality is moles per kg of solution (weight).

When dealing with dilute solutions there isn't normally a huge difference between them, but that gets larger as concentration increases.

As far as your question goes, what are you actually doing? You can't just use moles for pH as it is dependent on concentration, not total amount.

romanqwerty on
• edited November 2010
Okay heres an example:

A 1.0L buffers solution contains 0.100 mol HC2H3O & 0.100 NaC2H3O2. Ka = 1.8*10^-5. Initial pH is 4.74. Calculate pH after adding 0.010mol of solid NaOh.

OH- + HC2H3O2
>H2O + C2H3O2-
I 0.010mol 0.100mol 0.100mol
C -0.010mol -0.010mol +0.010mol
E 0 0.090mol 0.110mol

HC2H3O2 + H2O (EQ SIGN<>) H3O + C2H3O2-
I 0.090 0 0.110
C -x +x +x
E 0.090-x x 0.110+x

Ka=[H30][C2H3O2] use this to obtain x & thus H3O+ concentration
[HC2H302]

x=1.47*10^-5mol
pH=4.83

Otar on
[SIGPIC][/SIGPIC]
• my moons are good moons Registered User regular
edited November 2010
Otar wrote: »
Okay heres an example:

A 1.0L buffers solution contains 0.100 mol HC2H3O & 0.100 NaC2H3O2. Ka = 1.8*10^-5. Initial pH is 4.74. Calculate pH after adding 0.010mol of solid NaOh.

OH- + HC2H3O2
>H2O + C2H3O2-
I 0.010mol 0.100mol 0.100mol
C -0.010mol -0.010mol +0.010mol
E 0 0.090mol 0.110mol

HC2H3O2 + H2O (EQ SIGN<>) H3O + C2H3O2-
I 0.090 0 0.110
C -x +x +x
E 0.090-x x 0.110+x

Ka=[H30][C2H3O2] use this to obtain x & thus H3O+ concentration
[HC2H302]

x=1.47*10^-5mol
pH=4.83

At a quick glance, what you are doing is correct.
If you think about the solution, you are depleting base with acid or vise versa, the moles decreases, and is then adjusted by your total volume.

This will save you some time when you do titration calculations, so you're doing fine!!!!
Sorry DroppingLoads, but you are absolutely wrong about molality, and I'm not sure what you're talking about there. You might have been trying to compare formal solutions with molar solutions, but neither of those involve molality.

If that is the right answer in you're book, then you are doing fine. There are often multiple routes to a problem.

With a diprotic buffer like that, though, you would normally take pka1 and solve for all concentrations with help from the pka2 and your now known H+ concentration, but that is more an upper level problem.

Fuzzy Cumulonimbus Cloud on
• my moons are good moons Registered User regular
edited November 2010
you should also note that there is little change in the pH which means you are at least on the right track for a buffer solution

this is why buffers are so important
they have a huge capacity to resist change to pH in their buffer zone
(You'll learn why this is later, with help from Henderson and Hasslebach!)

Fuzzy Cumulonimbus Cloud on
• edited November 2010
Yeah, what you are doing is correct for a 1.0L solution. But remember when converting from x to pH that you have to take the logarithm of the concentration not the moles.

romanqwerty on
• my moons are good moons Registered User regular
edited November 2010
good catch

Fuzzy Cumulonimbus Cloud on
• edited November 2010
Sorry DroppingLoads, but you are absolutely wrong about molality, and I'm not sure what you're talking about there. You might have been trying to compare formal solutions with molar solutions, but neither of those involve molality.

You are absolutely correct, and thanks for the correction. I'm sorry about that Otar. For some reason I was thinking of fugacity, which applies to gasses.