quick chem pH question

[Sarahisme]

hey i can't figure this question out...

 

"What is the pH of the 1.0x10^-8 M HCL?"

 

could i please have some help? Smile

 

Thanks

 

Sarah

[H2SO4]

1-2

[BenSon]

-log (1x10^-8)

= pH of 8

 

~Scott

[Sarahisme]

my answers booklet says that its a bit less than 7 (~6.98)

[BenSon]

for a pH value of 6.98 the concentartion of acid would have to be (1 x 10^-6.98 M), I think, does the answer have working out?

 

~Scott

[Sarahisme]

The {H+}of neutral water (pH=7) is 1E-7. This HCl is a really weak solution (1E-8) of a strong acid, so you add the [H+] together (it's called the 5% rule and is used for weak acid solutions when the contribution of [H+] from the water is more than 5% of the [H+] from the acid), so in this case [H+]=1E-7+1E-8=1.1E-7.

 

And

 

pH=-log(1.1E-7)=6.96

 

This is a very weak acid solution.

[BenSon]

Alrighty what do you need us for then ?

 

~Scott

[Sarahisme]

well i didn't see how this worked really, but anyways

[neo007]

HCL are meant to be strong acids aren't they?

As far as i know they are, which means that most of the HCL dissociates into [H+] and [Cl-] ions. But maybe its because the concentration is so low that the pH is relatively high...

[BenSon]

well i didn't see how this worked really, but anyways

 

This is how I think it works neutral water 10^-7 for H+ and OH-. The HCL is 10^-8, If you plug that value into the standard pH formula you get pH 8 wich isn't realy true because you've added acid to neutral water it should be slightly acidic. By adding the the two amounts of H+ ions together you get the true pH. As for why it is only for 5% rule, My guess is that if the acid is stronger and the H+ from the water is less then 5% then the difference is negligable so we don't calculate it.

Example:

Ph of 1x10^-4 M HCl"

We plug that into the normal formula

-log(1x10^-4)

=pH of 4

If we do it the same as the 5% rule we get

-log((1 x 10^-4)+(1x10^-7)

=pH of 3.999956

 

~Scott

[Sarahisme]

ok thanks

[woelen]

hey i can't figure this question out...

 

"What is the pH of the 1.0x10^-8 M HCL?"

 

could i please have some help? Smile

 

Thanks

 

Sarah

 

For sake of brevity' date=' I introduce the symbol H for [H'] and OH for [OH].

 

For water we have H*OH = 10^-14

 

The acid HCl is a strong acid. At such low concentration, it can be assumed totally split.

 

Because for each OH(-) ion from water, we also have one H(+) ion, we now can say that H = OH + 10^-8, or OH = H - 10^8.

 

Now plug into the equation for water:

 

H * (H - 10^-8) = 10^-14.

 

H*H - 10^8 * H - 10^-14 = 0.

 

This is a quadratic equation in H. Solving this for H yields H = 1.05*10^-7.

Taking the -log10() of this yields 6.98, so the pH indeed equals 6.98.

 

The critical part of this computation is that at such very low concentrations, the contribution of the autoprotolysis of water cannot be neglected anymore.

[woelen]

The {H+}of neutral water (pH=7) is 1E-7. This HCl is a really weak solution (1E-8) of a strong acid' date=' so you add the [H+'] together (it's called the 5% rule and is used for weak acid solutions when the contribution of [H+] from the water is more than 5% of the [H+] from the acid), so in this case [H+]=1E-7+1E-8=1.1E-7.

 

And

 

pH=-log(1.1E-7)=6.96

 

This is a very weak acid solution.

What you are saying here is NOT correct! Your reasoning is not valid. You cannot simply add up the acid concentration to the concentration of 10^-7. Try this for [HCl] = 10^-7. With your reasoning you would have a concentration equal to 2*10^-7, but this is false.

 

The real concentration of H(+) will be 1.618*10^-7. I'll leave it as an exercise for you to explain this value.

[scientistsahai]

woelen that was good and logical. Most posts here don't really give any scientific answers.