How Do You Calculate pH After Adding NaOH to H2S Solution?

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Discussion Overview

The discussion revolves around calculating the pH of a solution after adding sodium hydroxide (NaOH) to a hydrogen sulfide (H2S) solution. Participants explore the implications of H2S being a weak acid and its dissociation steps, as well as the effects of adding a strong base. The context includes homework-related queries and attempts at solution methods.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a calculation method for determining pH after adding NaOH, arriving at a pH of 12.36, which they believe is incorrect based on expected results.
  • Another participant questions the understanding of the dissociation constants (Ka1 and Ka2) for H2S, noting it is a polyprotic acid.
  • Some participants clarify that H2S is a weak acid and discuss its dissociation in two steps, each with a different Ka value.
  • A participant suggests writing equilibrium equations for the dissociation reactions to find hydronium ion concentrations, indicating uncertainty about how this aids in the pH calculation.
  • There is a suggestion to use the Henderson-Hasselbalch equation with the second Ka to find the resulting pH after NaOH has been added.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the initial reasoning and calculations. Some agree on the need to consider the weak acid nature of H2S and its dissociation steps, while others challenge the initial calculations without reaching a consensus on the final pH value.

Contextual Notes

Participants note the complexity of calculating pH in a system involving a weak acid and a strong base, highlighting the need for careful consideration of equilibrium concentrations and the appropriate use of dissociation constants.

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Homework Statement



A vessel contains 500mL of .1 molar H2S solution. For H2S Ka1 = 1 x 10^-7
Ka2 = 1.3 x 10^-13.

What will the pH be when 800mL of .1molar NaOH has been added.

Homework Equations



ph = -log([H])
ph = pka + log(B/A)

The Attempt at a Solution


This is how I started, I found the moles of each. I found I started with .08 moles of NaOH and .05 moles of H2S, subtracted them to get .03 moles of NaOH. Then .03 / (.5 + .8) to get the concentration (total volume), then because all that's left is strong base I took the -log(.023) and then subtracted that from 14 to get the pH. I keep ending up with 12.36 but the answer is 13.1. I need to be able to do this for my exam tomarrow, can anyone help?
 
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What are k1 and k2?
 
Ka values, it is a polyprotic acid meaning its Hydrogens leave in steps and it has a different Ka value for each Hydrogen.
 
Well, I've not studied chemistry for quite a while now, but your reasoning seems fine to me. Sorry I can't be anymore help, but perhaps someone with a Chemistry background will read this and be able to help!
 
cristo said:
Well, I've not studied chemistry for quite a while now, but your reasoning seems fine to me. Sorry I can't be anymore help, but perhaps someone with a Chemistry background will read this and be able to help!
No - the reasoning is incorrect. H2S is not a strong base. It is a weak acid. Therefore it will disassociate in two steps, each one with a different Ka value.

Write an equation for equilibrium concentrations for the reaction which produces a hydronium ion and a deprotonated hydrogen sulfide. Then write another one which produces the hydronium ion and the sulfide ion. (Remember that in between these two reactions, the hydronium concentration is retained.)

It's a polyprotic disassociation, as you said.
 
H2S <==> HS + H

HS <==> H + S

I did this and I calculated the concentration of H at both steps
I don't know how this helps me though the concentration of the H ends up being around 1 x 10^-4.
 
physstudent1 said:

Homework Statement



A vessel contains 500mL of .1 molar H2S solution. For H2S Ka1 = 1 x 10^-7
Ka2 = 1.3 x 10^-13.

What will the pH be when 800mL of .1molar NaOH has been added.



Homework Equations



ph = -log([H])
ph = pka + log(B/A)


The Attempt at a Solution


This is how I started, I found the moles of each. I found I started with .08 moles of NaOH and .05 moles of H2S, subtracted them to get .03 moles of NaOH. Then .03 / (.5 + .8) to get the concentration (total volume), then because all that's left is strong base I took the -log(.023) and then subtracted that from 14 to get the pH. I keep ending up with 12.36 but the answer is 13.1. I need to be able to do this for my exam tomarrow, can anyone help?

You have .1 mole equivalent amount of acid and .08 moles of NaOH, in relative terms, which mean that after the NaOH has been consumed, you're going to be left with concentrations of S 2- and HS-...use the Henderson Hasselbach equation using the second Ka to find the pH of the resulting solution.
 

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