# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Sophia7X on October 20, 2012, 07:05:31 PM

Title: couple questions {pH}
Post by: Sophia7X on October 20, 2012, 07:05:31 PM
What is the pH of a 5e-8 M HCl solution?

My attempt at solution:
I don't think you can -log that because that gives you a basic pH and adding a strong to water however little in amount shouldn't make pH basic.
ionization of water: H2O  ::equil:: H+ + OH-
Difference between H+ and OH- concentration must be 5e-8 since HCl is a strong acid so assume 100% dissociation into H+.
[H+] - [OH-] = 5e-8 ([OH-] = -5e-8+[H+])
Kw = 1e-14 = [H+][OH-]

I substituted [OH-] into Kw, 1e-14 = [H+][-5e-8+ [H+]]
[H+] = 1.28e-7
pH = 6.89 but apparently the answer is 7.00... how can it be completely neutral?

________________________________________________________________

A + B --> AB is 1st order with respect to A, zero order with respect to B. Initial concentration of A and B = both 0.100 M. After 1.5 hours, conc of B = 0.060 M. What is the specific rate constant?

ln 0.06 = -k(1.5 hr) + ln 0.100
k = 0.34 1/hr
but the answer is 0.61 1/hr, how do I get this?
Title: Re: couple questions {pH}
Post by: curiouscat on October 21, 2012, 12:05:50 AM
A + B --> AB is 1st order with respect to A, zero order with respect to B. Initial concentration of A and B = both 0.100 M. After 1.5 hours, conc of B = 0.060 M. What is the specific rate constant?

ln 0.06 = -k(1.5 hr) + ln 0.100
k = 0.34 1/hr
but the answer is 0.61 1/hr, how do I get this?

I get 0.34 hr-1 too.
Title: Re: couple questions {pH}
Post by: Borek on October 21, 2012, 05:29:46 AM
pH = 6.89

6.89 is a correct answer.
Title: Re: couple questions {pH}
Post by: Sophia7X on October 22, 2012, 02:11:53 PM
pH = 6.89

6.89 is a correct answer.

Thanks

ooh that's a cool program
Title: Re: couple questions {pH}
Post by: Rutherford on October 23, 2012, 07:26:31 AM
A + B --> AB is 1st order with respect to A, zero order with respect to B. Initial concentration of A and B = both 0.100 M. After 1.5 hours, conc of B = 0.060 M. What is the specific rate constant?

ln 0.06 = -k(1.5 hr) + ln 0.100
k = 0.34 1/hr
but the answer is 0.61 1/hr, how do I get this?
I got their answer when replacing ln0.06 with ln0.04 but 0.04 is the concentration change, not the amount that left (which is put in the rate law). I can't tell right away if it is a mistake or not.
Title: Re: couple questions {pH}
Post by: curiouscat on October 23, 2012, 07:40:33 AM
I got their answer when replacing ln0.06 with ln0.04 but 0.04 is the concentration change, not the amount that left (which is put in the rate law). I can't tell right away if it is a mistake or not.

Unless the question meant to say: "After 1.5 hours, conc of B AB = 0.060 M"
Title: Re: couple questions {pH}
Post by: Aegis6 on October 24, 2012, 05:57:32 PM
I got a pH of 7.3

My math:

pH = -log [H+]

5x10-8 M HCl = 5x10-8 M H+

Once again, pH = -log [H+]

-log(5x10-8) = 7.3

Using sig figs, 7.3 becomes just 7.

QED, pH=7

Wolframalpha for verification: http://www.wolframalpha.com/input/?i=pH+5x10%5E-8+M+HCl
Title: Re: couple questions {pH}
Post by: Rutherford on October 25, 2012, 08:21:58 AM
At such low concentrations water dissociation takes an important role. Sophia wrote it nicely in the first post.

For the second one, it has to be the concentration change of AB, because the concentrations of B and A aren't proportional and they can't be compared.
Title: Re: couple questions {pH}
Post by: curiouscat on October 25, 2012, 08:25:19 AM
For the second one, it has to be the concentration change of AB, because the concentrations of B and A aren't proportional and they can't be compared.

You are confusing stoichiomety and kinetics. Conc. of A and B must remain equal at all times.
Title: Re: couple questions {pH}
Post by: Borek on October 25, 2012, 08:47:29 AM
I got a pH of 7.3

So the solution of an acid is slightly basic?

Quote
Wolframalpha for verification: http://www.wolframalpha.com/input/?i=pH+5x10%5E-8+M+HCl

Several years ago WA was told their results are wrong - and they were told it many times, on many occasions. They slightly modified the engine so that when the result becomes nonsensical it is displayed as 7. Highly disappointing.
Title: Re: couple questions {pH}
Post by: Aegis6 on October 25, 2012, 06:41:49 PM
I got a pH of 7.3

So the solution of an acid is slightly basic?

The method for determining the pH of a strong acid or base is to take the negative logarithm of the concentration of H+ ions.

As you have 5x10-8 M HCl, and you have one mol of H+ per mole of HCl, it logically follows that you have 5x10-8 M H+.

pH = -log[H+]
pH = -log[5x10-8]
pH = 7.30

I couldn't possibly tell you why the value is basic, but that is the value you get when you calculate for the pH of the HCl solution.
Title: Re: couple questions {pH}
Post by: Borek on October 26, 2012, 04:16:50 AM
As you have 5x10-8 M HCl, and you have one mol of H+ per mole of HCl, it logically follows that you have 5x10-8 M H+.

No, it doesn't follow. Sophia explained the correct approach in the very first post of this thread.
Title: Re: couple questions {pH}
Post by: Rutherford on October 26, 2012, 09:28:41 AM
For the second one, it has to be the concentration change of AB, because the concentrations of B and A aren't proportional and they can't be compared.

You are confusing stoichiomety and kinetics. Conc. of A and B must remain equal at all times.
How do you mean?
The rate law for the concentration of A is: ln[A]=ln[A]0-kt
For B it is: [B0]=[B0]0-kt
After time t, they can't be equal (added 0 above B because without it I get bold letters).
Title: Re: couple questions {pH}
Post by: curiouscat on October 26, 2012, 01:42:42 PM
How do you mean?
The rate law for the concentration of A is: ln[A]=ln[A]0-kt
For B it is: [B0]=[B0]0-kt
After time t, they can't be equal (added 0 above B because without it I get bold letters).

Your rate expressions are wrong.

To make the fallacy evident: Assume you had 1 mol of both A and B to start with. If at some later time if you end up with (say) 0.4 mol A and 0.5 mol B that means more of A reacted than B.  Correct?

What did that excess A react with if not B?  If A+B-->C then dA=dB at all t.

Makes sense?
Title: Re: couple questions {pH}
Post by: Sophia7X on October 26, 2012, 10:42:29 PM

The method for determining the pH of a strong acid or base is to take the negative logarithm of the concentration of H+ ions.

Hmm, not when the strong acid's concentration is less than the ~1*10-7 since at such a dilute concentration, ionization of water will play a role

I got their answer when replacing ln0.06 with ln0.04 but 0.04 is the concentration change, not the amount that left (which is put in the rate law). I can't tell right away if it is a mistake or not.

In that case, I guess the question just had a typo.
Title: Re: couple questions {pH}
Post by: Rutherford on October 27, 2012, 04:06:18 AM
Isn't it the same situation like here:
2A+B :rarrow: C
A is second order, B is first order. When 1 mole of B disappears, 2 moles of A disappear. If a reactant is zero order, his concentration doesn't affect the rate until its c reaches zero.

"What did that excess A react with if not B?  If A+B-->C then dA=dB at all t."
To answer that, we should know the mechanism which is not given, so without it we are just going in a cycle.
Title: Re: couple questions {pH}
Post by: Borek on October 27, 2012, 04:40:31 AM
Why do you ignore stoichiometry? Reaction has to follow both the stoichiometry and laws of kinetics, it can't follow one of them but ignore the other.

If the reaction equation is A + B :rarrow: C no matter what kinetics is, 1 mole of A always react with 1 mole of B. If 1 mole of A disappears, 1 of mole of B disappears. There is no way that something else can happen, no matter what is the mechanism.
Title: Re: couple questions {pH}
Post by: Rutherford on October 27, 2012, 05:29:15 AM
I see now that I got confused, because I saw the different plots for different order reactants. Now I understood that the concentration change of B depends on the concentration change of A (they must have the same plot regardless of the order).