What Is the Correct pH of a 0.15-M NaX Solution?

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Discussion Overview

The discussion centers around determining the pH of a 0.15-M aqueous solution of NaX, a salt derived from a weak acid (HX) and a strong base (NaOH). Participants explore the implications of hydrolysis and the relationship between the acid dissociation constant (K(a)) and the hydrolysis constant (K(h)) in calculating pH.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the H+ concentration as 1.06 x 10-6 and derives a pH of 5.97, but expresses uncertainty about the next steps.
  • Another participant asserts that NaX will be basic due to hydrolysis, suggesting that the pH must be above 7 and that the hydrolysis equilibrium should be considered.
  • A participant emphasizes the need to find the hydrolysis constant Kh in terms of Ka and Kw to proceed with the calculation.
  • There is a discussion about the utility of a pH cheat sheet, with some participants arguing that it may not be helpful without a proper understanding of the underlying concepts.
  • Concerns are raised about encouraging reliance on formulas without comprehension, with a preference for deriving equations to foster understanding.

Areas of Agreement / Disagreement

Participants express differing views on the use of the cheat sheet and the importance of understanding the underlying principles before applying formulas. There is no consensus on the best approach to calculating the pH of the NaX solution.

Contextual Notes

Participants highlight the need to consider hydrolysis and the relationship between Ka and Kw, but specific mathematical steps and assumptions remain unresolved.

confusedbyphysics
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K(a) for HX is 7.5 x 10^-12. What is the pH of a 0.15-M aqueous solution of NaX?

a. 7.87
b. 1.85
c. 5.97
d. 8.03
e. 12.15


HX --> H+ + X-

I found H+ concentration, which is 1.06 X 10-6, then found pH of this, and it is 5.97. I wasn't sure where to go from here so I guessed 5.97 and it is wrong (I get another guess).

How do I go from finding the pH of the HX to finding the pH of the NaX? 14 - 5.97 - 8.03 which is an answer, but I'm not sure why I'd do this? isn't the conjugate base just the X-? Thanks for any help
 
Last edited:
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First of all, since NaX is the salt of a strong base and weak acid, the salt will be basic on hydrolysis. So you can eliminate all answers below 7.

HX --> H+ + X-
I found H+ concentration, which is 1.06 X 10-6, then found pH of this, and it is 5.97. I wasn't sure where to go from here so I guessed 5.97 and it is wrong (I get another guess).

That's wrong, because that's not the hydrolysis equilibrium for the salt. What happens is that the X- will react with water to form HX and OH-.
You will first need to find the Hydrolysis constant Kh in terms of Ka and Kw.
Then write down the relation between the concentration of the species at equilibrium. Since you know Kh (becaues you know Ka and Kw), and the initial concentration, you will be able to find the concentration of OH- and hence the pH.
 
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Check out this pH cheat sheet. But you can't use it before you will understand what you are doing :smile:
 
Borek, that cheat sheet isn't very useful if the OP doesn't actually understand what's happening. In fact, it would encourage people to blindly apply a formula without actually learning the concepts.

The sheet will be useful and will save time only for experts like you, who have already derived it by hand and know what's happening.
 
siddharth said:
The sheet will be useful and will save time only for experts like you, who have already derived it by hand and know what's happening.

Well, it seems to me you know enough to use it :smile:
 
Borek said:
Well, it seems to me you know enough to use it :smile:

Yeah, but does the original poster know?

I wanted confusedbyphysics to actually derive that equation by himself/herself, instead of looking up the final formula, because then confusedbyphysics would have understood some of the concepts involved.
 
Well, I stated "you can't use it before you will understand what you are doing". So if he tries - he does it on his own risk :wink:

Sorry for interfering with your pedagogical plan :smile:
 

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