Calculating the Ka from pH and initial concentration

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Discussion Overview

The discussion centers around calculating the acid dissociation constant (Ka) from the pH and initial concentration of a weak acid solution. Participants explore the calculations involved in determining percent ionization and the correct value of Ka, addressing potential errors in computation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the hydronium ion concentration from pH and finds it to be 4.5*10^-3, leading to a percent ionization of 22.5%.
  • The same participant attempts to calculate Ka using the equilibrium expression and arrives at a value of 1.3*10^-3, questioning the correctness of this result based on feedback from their professor.
  • Another participant suggests that the Ka value is incorrect due to a miscomputation and encourages a re-evaluation of the numeric expression.
  • A different participant expresses confidence in their repeated calculations, asserting they consistently arrive at the same Ka value.
  • Another reply advises checking the computation manually, implying that the calculator may be misused, and hints that the expected Ka value should be significantly smaller.
  • One participant confirms the initial calculations for hydronium concentration and percent ionization but disputes the need to solve for 'x' in the equilibrium expression, suggesting a misunderstanding in the approach.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correctness of the calculated Ka value, with some asserting it is incorrect while others maintain their calculations. The discussion remains unresolved as no consensus is reached on the correct value of Ka.

Contextual Notes

There are indications of potential computational errors, particularly in the use of calculators, and some confusion regarding the equilibrium expression and the need for solving for 'x'.

maceng7
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Homework Statement


A 0.0200 mol/L acid solution has a pH of 2.347. Calculate the percent ionization of this acid. Calculate the Ka of this acid

The Attempt at a Solution



I first calculated the hydronium ion concentration: 10^-pH = 10^-2.347 = 4.5*10^-3

I then used this concentration to calculate the percent ionization: 0.0200 / 4.5*10^-3 = 22.5%

Can anyone confirm this answer. Now the second part of the question asks for the Ka. My prof said that if you got 1.3*10^-3 as the Ka value it is wrong. I keep getting this value and it makes no sense why it would be anything other than this value:

Since the ratio in the balanced chemical equation is 1:1 , the equilibrium expression is:
Ka = (4.5*10^-3)(4.5*10^-3) / (0.0200 - 4.5*10^-3)

solving for x I get 1.3*10^-3, however this isn't suppose to be the right answer.. can anyone tell me where I've gone wrong or maybe my prof is mistaken? Thanks.
 
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Your Ka value is wrong simply because you miscomputed your numeric expression. The numeric expression seems good; Try recomputing and you will, or should find a much smaller Ka value.
 
I don't know I've put it into my calculator numerous times and I get the same value,
 
You are incorrectly handling this computation on the calculator. Your expression seems to be written well and appears credible, so as a check, try doing the computation the old fashioned, manual way. You SHOULD find something about ?.?? x 10-6
 
maceng7 said:
I first calculated the hydronium ion concentration: 10^-pH = 10^-2.347 = 4.5*10^-3

OK

I then used this concentration to calculate the percent ionization: 0.0200 / 4.5*10^-3 = 22.5%

Can anyone confirm this answer.

No.

\frac {0.0200} {4.5\times 10^{-3}} = 4.44

Since the ratio in the balanced chemical equation is 1:1 , the equilibrium expression is:
Ka = (4.5*10^-3)(4.5*10^-3) / (0.0200 - 4.5*10^-3)

solving for x I get 1.3*10^-3

Expression looks OK to me, but first - there is no x to solve for, second - as symbolipoint stated, it evaluates to something completely different.
 

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