What is the Hydroxide Ion Concentration in an Ionic Equilibrium Problem?

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Discussion Overview

The discussion revolves around calculating the hydroxide ion concentration in a reaction between sulfuric acid and sodium hydroxide, focusing on stoichiometry and ionic equilibrium. Participants explore different approaches to the calculation and the implications of the reaction equation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation of hydroxide ion concentration based on the volumes and molarities of sulfuric acid and sodium hydroxide.
  • Another participant critiques the initial approach, suggesting that the problem is more about stoichiometry than equilibrium.
  • A different participant attempts a revised calculation, incorporating the stoichiometric coefficients from the balanced reaction.
  • Concerns are raised about the accuracy of multiplying by the coefficient of NaOH without proper justification.
  • Further clarification is provided regarding the number of moles of acid and the corresponding amount of NaOH that reacts.
  • A final calculation is presented, showing the excess hydroxide ions after accounting for the reaction with sulfuric acid.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to the problem, with some emphasizing stoichiometry and others focusing on the equilibrium aspect. No consensus is reached on the final calculation or methodology.

Contextual Notes

Participants highlight the importance of correctly applying stoichiometric coefficients and the need to account for the moles of reactants involved in the reaction. There are unresolved questions about the accuracy of the calculations presented.

Speedking96
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Homework Statement



Calculate the hydroxide ion concentration:

16.5 mL of aqueous sulfuric acid at 1.5 M added to 12.7 mL of sodium hydroxide at 5.5M.

2. The attempt at a solution

H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

Hydrogen Concentration: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
Hydroxide Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles

0.0451 moles of Hydroxide / (16.5 + 12.7 mL) = 1.54 M
 
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This is hardly an equilibrium, looks like a simple stoichiometry to me.

You have wrote the reaction equation (good idea) but then you ignored it (bad idea).
 
H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

Hydrogen Concentration: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
Hydroxide Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles x 2 = 0.1397

0.0.11495 moles of Hydroxide / (16.5 + 12.7 mL) = 3.93 M
 
With some luck third guess will be the correct one.
 
Why? I incorporated the NaOH coefficient.
 
Just because you multiplied something by 2 doesn't mean you did it correctly.

How many moles of acid do you have? How many moles of NaOH will react with this acid?
 
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H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

Acid: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
NaOH Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles

0.0495 moles of NaOH will react with the acid.

Extra OH: 0.06985 - 0.0495 = 0.02035 moles

Concentration: (0.02035) / (16.5 + 12.7 mL) = 0.7 M
 
Looks OK now.
 

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