Calculating pH in Geometric Units: Solving a Tricky Acidic Battery Question

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To determine the pH of a battery acid that is one-fifth as acidic as it should be (with a target pH of 0.8), one must first understand that "one-fifth as acidic" relates to the concentration of hydrogen ions, not a direct division of the pH value. The correct approach involves calculating the concentration of H+ ions from the original pH, then dividing that concentration by five, and finally converting back to pH using the formula pH = -log[H+]. The calculations yield a pH of approximately 1.50 for the less acidic solution. It's important to note that pH values are typically reported with only two or three significant figures for practical purposes.
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Homework Statement



If the ph of battery acid should be 0.8, what is the pH of the acid in a battery that is one-fifth as acidic as it should be?

Homework Equations





The Attempt at a Solution


We just learned how to comparing pH in Geometric unit, using the formula 10^(ph1-ph2).
But for this question, do we really need to use that formula, or we just 0.8/5=0.16?
ty!
 
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Suy said:

Homework Statement



If the ph of battery acid should be 0.8, what is the pH of the acid in a battery that is one-fifth as acidic as it should be?

Homework Equations





The Attempt at a Solution


We just learned how to comparing pH in Geometric unit, using the formula 10^(ph1-ph2).
But for this question, do we really need to use that formula, or we just 0.8/5=0.16?
ty!

How is this a geometric question?

To answer your question, you need to find out what "one-fifth as acidic" means. I suspect that it means that the concentration of H ions is one-fifth what the concentration it should be, but that doesn't mean that the pH will be one-fifth of what it should be.

10^(pH1-pH2) is not a formula, since it is not an equation. Look up the equation you need.
 
"one fifth as acidic" seems to be directly related to concentration, not pH of concentration. You would focus on concentration of hydronium ions.
 
so it would be 1/5=10^(0.8-ph2)?
Am i right?
 
Suy said:
so it would be 1/5=10^(0.8-ph2)?
Am i right?

No. You need to convert directly to concentration, perform your specified calculation, then convert back to pH.
 
symbolipoint said:
No. You need to convert directly to concentration, perform your specified calculation, then convert back to pH.
But isn't they are the same?
pH=-log[H+]
-0.8=log[H+]
10^(-0.8)=H+
H+=0.15848931924611134852021013733915

0.15848931924611134852021013733915/5
=0.03169786384922226970404202746783

-log[0.03169786384922226970404202746783]
=1.50
----------------------------------------------
The book teach us to do it this way
10^(ph1-ph2)
10^(0.8-ph2)=1/5
ph2=1.50

anyway.., is this right?
ty!
 
I myself don't follow what your book is teaching but neither am I trying to analyze it. No matter - just use fundamental principles and you should be able to think your way through effective steps.

One thing you should know is that your use of excessive accuracy does not give you better numeric value information. Usually, pH's are good to only about 2 or 3 significant figures.
 
symbolipoint said:
I myself don't follow what your book is teaching but neither am I trying to analyze it. No matter - just use fundamental principles and you should be able to think your way through effective steps.

One thing you should know is that your use of excessive accuracy does not give you better numeric value information. Usually, pH's are good to only about 2 or 3 significant figures.

oh, because i use the computer calculator. I just copy the number and put it here. I won't do that in my assignment.
Thx for helping me!
 
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