# calculating pH given volume and molarity

Hey,

If anyone could help me with these questions I would really appreciate it!

1. Titrate 5 mL of 1.5 M NaOH into 25 mL of 2.0 M HCl. Find pH.

2. Find the pH of .1 M H2SO4 knowing that Ka1=strong acid and Ka2= 1.2 x 10^-2

3. Find pH of 10 micromolar HIO (a weak acid), ka=2.3 x 10^-11

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Hello,

You atleast need to work out something, and if you go wrong we are there to help you out. but first atleast show us some of your work.

I hope you understand as we want to make you understand the question on how to solve rather than solving it entirely for you.

so pls. try to work out something, show us your work and we shall help you with it.

hope that helps.

Anonymous (not verified)

Tue, 2008-03-11 08:56

Oh, sorry I didn't realize that's how this worked. Trust me I've spent a lot of time on these problems!

1. I determined this was before the equivalence point (7.5 mol NaOH and 50 mol HCl). Then found that after all the OH was consumed there would be 42.5 mol HCl remaining (so concentration of H would be 1416.67. Took - log and got a negative pH so obviously I did something wrong.

2 Reasoned that it would dissociated first into HSO4 and H and since its strong it would dissociate completely and yield .1 M H. Then calculated 1.2 x10^-2 x (SO4-2)(H)/.1M and tried to add the two concentrations of H together and get pH from that. Again, didn't work.

3. Set up Ka like above and calculated pH from concentration of H but pH was larger than 7 and HIO is a weak acid.

Where am I going wrong in these calculations?

lauralaura

Tue, 2008-03-11 13:52

well I at least tried the 10 uM one, but I didn't get the right answer, so help would be appreciated:

Ka= X

^{2}/ (1*10^{-5}-x)= 2.3*10^{-11}X

^{2}=2.3*10^{-11}* -(X)(1*10^{-5})0=-X

^{2}-2.3*10^{-11}(X)+2.3*10^{-16}using the quadratic formula gives X=1.515*10

^{-8}pH=-log(1.515*10

^{-8})=7.82but it is an acid, so I know it should be

catalina

Tue, 2008-03-11 14:01

1. I determined this was before the equivalence point (7.5 mol NaOH and 50 mol HCl). Then found that after all the OH was consumed there would be 42.5 mol HCl remaining (so concentration of H would be 1416.67. Took - log and got a negative pH so obviously I did something wrong.

I think that your mole calculations are incorrect: moles of sample = Molarity x volume (in liters)

For NaOH = 1.5 mol/L x .005L = .0075 mol NaOH

For HCl = 2.0 mol/L x .025L = .050 mol HCl

Moles of excess HC = .050 - .0075 = .0425 mol

Molarity of excess HCl solution = .0425 mol / .030 L = 1.41 M

Since each HCl molecule produces 1 H+ ion, Molarity of H+ = 1.41 M

pH = - log (1.41) = -.146

Arriving at a negative pH made me uncomfortable, since that is not the usual situation. However, I checked the following site and found out that it is possible for pH in concentrated acid solutions (and 1.41 M is pretty concentrated) to be negative. http://www.newton.dep.anl.gov/askasci/chem99/chem99230.htm

2. Find the pH of .1 M H2SO4 knowing that Ka1=strong acid and Ka2= 1.2 x 10^-2

Since Ka1 is given as "strong acid, we can assume that the first ionization goes to completion:

H2SO4 --> H+ + HSO4 -

.1 M .1M .1M

The second ionization will NOT go to completion, but will be an equilibrium situation. Let's do an ICE Table:

HSO4- H+ + SO4-2

Initial .1 .1 0

Change -x +x +x

Equilibrium .1 - x .1 +x x

Plug into the Ka2 expression:

Ka2 = [H+][SO4-2]/ [HSO4-]

to get

Ka2 = (.1 +x) (x) / (.1 - x)

Solve for x and then find the pH.

3. Find pH of 10 micromolar HIO (a weak acid), ka=2.3 x 10^-11

10 micromolar HIO = 10 x 10-6 M or 1 x 10-5 M Let's do another ICE Table:

HIO H+ + IO-

Initial 1E-5 1E-7 0 ( [H+] in pure water = 1E-7)

Change -x +x +x

Equilibrium 1E-5 -x 1E-7+x x

Due to the size of the Ka value, you can assume 1E-5 - x = 1E-5 and 1E-7 + x will also be approximately 1E-7.

(Or you can use the quadratic formula to get an exact solution)

The acid is so weak that it will not appreciably change the [H+]. pH of resulting solution will be 7.

spock

Tue, 2008-03-11 15:35

I understand the first two questions, but on the third, why is the initial concentration of H+ = concentration of H+ in water? that is the only thing that is different from how I originally did it.

also, when I use the quadratic equation to try and get an exact answer, i get a pH=8.648. If anything other than 7, shouldn't the pH be a little less?

Making the assumptions, I calculate a pH=7.819. Is this close enough to 7 to still be considered neutral?

catalina

Tue, 2008-03-11 16:25

Well, if you ignore the presence of water the [H+] comes back with a value of 1.41 x 10-8 for a pH of 7.85. That doesn't make sense, we take an acid and add it to pure water and the solution becomes basic!?!?!

That indicates that the [H+] would be less than pure water. That cannot be correct if we are starting with pure water (that already has a [H+] of 1 x 10-7) and then add additional H+ ion to it from the ionization of the acid. We would expect the H+ concentration to increase, or at least not change SIGNIFICANTLY.

In the first two cases, you could put in 1E-7 for the initial H+ concentration as well, but in those cases the amount of additional H+ ion added is much greater than 1E-7 making the initial concentration of H+ insignificant. That is the case with most acids, but the weaker the acid, the more significant the initial concentration of H+ becomes.

In the third case, the initial concentration of H+ ion in pure water cannot be ignored because it IS significant (in fact it is larger than the amount of H+ added by the acid).

I did it again without using the assumptions that 1E-7+x and 1E-5 - x were equal to 1E-7 and 1E-5 and used the quadratic formula to solve for x.

I came up with a value of 2.25E-9 for x

Given that [H+] is (1E-7 + x) then the value resulting from the exact treatment in the quadratic formula predicts a H+ concentration of 1.0225 x 10-7M

and a pH of 6.99. I think that verifies the original assumptions which led me to conclude that the pH of the solution would be 7.

spock

Tue, 2008-03-11 16:58