pH question

[thais]

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Can someone help me out with this question:

If 99% of the H+ ions in a solution of pH 1.3 were neutralized, what would be the final pH?

The answer is 3.3.

I don't really understand why.
Thank you!

 

[chiddler]

beginning pH 1.3 = 10^-1.3 H ions = 5*10^-2

so you reduce it by 99% - or take 1% of that

0.01 * 5*10^-2 = 5 * 10^-4

that number is the new H+ concentration. so we find pH with that concentration

pH = -log (5*10^-4) = 3.3

 

[SLC]

Short answer is that for a full 1 point increase in pH, you have to decrease your [H+] by a factor of 10.

In this case we're decreasing the [H+] by a factor of 100 (essentially) so we raise the pH by 2 full points (1.3 to 3.3)

 

[thais]

And how do you know that H ions = 5*10^-2 ?

I know I've learned that Kaplan method of subtracting the multiplier as a decimal number from the exponent.
2 - 0.5 ~ 1.5
Is this the method you use?

 

[chiddler]

i use this method

 

[thais]

THANK YOU guys! 🙂

 

[SaintJude]

Short answer is that for a full 1 point increase in pH, you have to decrease your [H+] by a factor of 10.

In this case we're decreasing the [H+] by a factor of 100 (essentially) so we raise the pH by 2 full points (1.3 to 3.3)

Ah, thanks for this insight. That's perfect MCAT strategy.

 

[MedPR]

Short answer is that for a full 1 point increase in pH, you have to decrease your [H+] by a factor of 10.

In this case we're decreasing the [H+] by a factor of 100 (essentially) so we raise the pH by 2 full points (1.3 to 3.3)

While Chiddler's method is good for demonstrating impressive log skills, I think this one is a tad bit better for MCAT 🙂

 

[MedPR]

beginning pH 1.3 = 10^-1.3 H ions = 5*10^-2

I'm not following how you got there either. That link you gave didn't explain how to handle fractional exponents, only fractional coefficients.

 

[chiddler]

huh.

i don't know how i got that. i mean i know it was correct, but i think it was just memorized or something...

lol. i hope my mcat is like this too xD

 

[MedPR]

huh.

i don't know how i got that. i mean i know it was correct, but i think it was just memorized or something...

lol. i hope my mcat is like this too xD

Haha so you're saying you have 10^-1.3 = 5*10^-2 memorized? Maybe after all this I will have it memorized too.. 🙂

 

[gentlebeast]

using HH is easier
pH= pKa+ log([base]/[acid])
we don't know pKa here, but it is safe to use the original pH=1.3
therefore pH=1.3+log(99/1)=1.3+2=3.3

 

[MedPR]

using HH is easier
pH= pKa+ log([base]/[acid])
we don't know pKa here, but it is safe to use the original pH=1.3
therefore pH=1.3+log(99/1)=1.3+2=3.3

Why is it safe to set pkA=pH?

 

[RiPSTONE]

And how do you know that H ions = 5*10^-2 ?

I know I've learned that Kaplan method of subtracting the multiplier as a decimal number from the exponent.
2 - 0.5 ~ 1.5
Is this the method you use?

I'm also curious as to how you got the 5*10^-2 ions. Do you mind explaining that step by step?

 

[Czarcasm]

I'm also curious as to how you got the 5*10^-2 ions. Do you mind explaining that step by step?

pH of 1.3 is somewhere in between 10^-1 and 10^-2 [H+]. Because it's a logarithmic scale , 5x10^-2 is approximately equal to a pH of 1.3. From there, knowing 99% neutralized, this essentially means you have 1/100 H+ left. 1/100 of 5x10^-2 [H+] is 5x10^-4 or approximately, pH of 3.3.

Technically, you didn't even need to do the conversion if you just realized that a decrease by a factor of 100 is equivalent to 2 pH units. In this case, because they are inversely proportional, it would be an increase of 2 pH units, hence 3.3.