[newgrad1]

Hello all!

This is a concept that has been bugging the hell out of me.. can somebody put me out of my misery with the calculations to go about solving it? I know I am missing a piece of the calculation puzzle.

I have 10 L of pure water at pH7 and I have 2M HCl solution. How much acid would I have to add to the water to lower the pH to 2?

Thanks

[sjb]

Any thoughts of your own? For instance, what concentration of H⁺ are you now aiming for?

[newgrad1]

Well you are now aiming for a 1 x 10^-2M of H⁺

This represents a difference of 9.99 x 10^-3M in comparison to the original pH7 of the water.

[sjb]

So, for instance if you add 10 ml of your HCl, what is the new concentration of acid?

[newgrad1]

Ah OK thanks.... so there is a different approach required than I had envisaged! I am going with adding 1 ml rather than 10 ml for the additional accuracy in the calculations??

Well for every 1 ml of 2M HCl added:

Adds 2 x 10^-3 mols
Current mol = 2.001 x 10^-3 mols

Current [H⁺] = 2.0008 x 10^-4 M
Represents an addition of 1.9997 x 10^-4 M

Δ[H⁺] required  /  Δ[H⁺] per 1 ml of HCl added

= 9.99 x 10^-3 M  /  1.9997 x 10^-4 M

= 49.957

= 49.96 ml of 2M HCl needed (addition of 9.99 x 10^-2 mols)

Check Final [H^+]]:

(1 x 10^-6 mols) + (9.99 x 10^-2 mols)  /  10 L + 0.04996 L

= 9.94 x 10^-3 M

pH = -log[H⁺]
= 2

I think where I was going wrong was that I was trying to formulate an equation to just plug the figures into rather than use a more stepwise approach? What do you think?

[Borek]

You can do it in one step, no problem at all, just express the H⁺ as a function of one unknown (V_added) and solve. HS algebra.

On the practical side: you can safely ignore the initial amount of the H⁺ present, it is several orders of magnitude lower than the final, and you won't be able to measure volume with accuracy high enough to take this difference into account (measuring 1±0.01 mL is a lab standard, measuring 1±0.0001 mL is quite a challenge).

On the theoretical side: H⁺ comes from the water autodissociation, this is an equilibrium reaction, and the equilibrium will shift once the pH goes down, so the amount of H⁺ from the water autodissociation at pH 2 will be different from the amount present initially.

[newgrad1]

You can do it in one step, no problem at all, just express the H⁺ as a function of one unknown (V_added) and solve. HS algebra.

I tried doing similar to this before but there was something wrong in my maths....

Final #mols HCl - Initial #mols HCl = #mols HCl added

(10 + y) x 10^-2 - (10 x 10^-7) = y9  (where y = volume of HCl added)

Solve for y

The final answer came out looking way off??

[Borek]

What is y9 intended to mean?

The general idea looks OK (even if - as I wrote earlier - it is a bit off, as the initial amount of H⁺ is not what you assume it is, but the difference it makes is orders of magnitude too low to matter).

[newgrad1]

Sorry typo.... should have read 2y (ie 2M x unknown volume = mols added)

Looking at the calculations again, yes I see what you mean about ignoring the initial [H⁺]

(10 + y) x 10^-2 = 2y

10^-1 + 10^-2y = 2y

10^-1 = 1.99y

y = 0.050 L   (very close to what the longer calculation turned out to be)


Regarding your comment about ignoring initial [H⁺], presumably if you were talking about a change from pH2 to pH0, then you would factor in the initial conc?

[AWK]

At pH = 2 (c_H3O⁺ = 10^-2), the concentration of H₃O⁺ ions from water dissociation is 10^-12. The sense of taking this second value into account would only be when the calculations should be carried out with an accuracy of more than 10 significant digits, and you use a maximum of 3 significant digits.

[newgrad1]

OK yes that makes sense....thanks to all concerned!