[vort3x]

Hi,

I have a bit of a silly question but just wanted to check. I have to write an equation for when this substance acts as an acid with water (therefore donating a proton), but does it matter whether I decrease H₂ or H₇ in the forward reaction? I'm pretty sure it doesn't since both are counted towards the overall hydrogen count.

H₂C₆H₇O_5(aq) + H₂O_(l) HC₆H₇O₅^-_(aq) + H₃O⁺_(aq)

Thanks!

[Astrokel]

Nope if you notice why the acid is not written as C₆H₉O₅ instead because what they are telling you it is an diprotic acid, so what you did is correct - you should not decrease H₇ instead.

[vort3x]

Another question. Now I'm calculating the pH of strong acids (-log of molarity). Which is fine, but in this question it asks,

pH of: 5.00ml of 1.00M HCl is diluted to 0.500L.

So using the dilution equation (M1V1 = M2V2), I got a final molarity of 0.01M and therefore the pH is 2. Can anyone confirm? (just want to make sure I'm approaching these right).

[Borek]

OK

[vort3x]

Ok, thanks.

Next question:
Calculate pH and [OH^-] of solution:

10.0ml of 0.015M Ba(OH)₂ with 40.0ml of 7.5x10^-3M NaOH.

I just did a bunch of similar questions where I was given just molarity, but not sure how to approach it with that solution, to get a "final" molarity to calculate pH from?

[Astrokel]

What is the total moles of OH^- and what is the volume of the final solution?

[vort3x]

The total moles of OH^- are 2, and the final volume is 50.0 ml = 0.05L. Still not sure.

Do I re-find molarity for each but just using the new volume? Ex:

Moles of Ba(OH)₂ / 0.05L
Moles of NaOH / 0.05L

Then add the two molarities for my final molarity?

... wait I guess I couldn't do that since I don't have a mass, hmmm.

[Borek]

Sum moles of OH^- and divide by the final volume.

[vort3x]

So 3 mol of OH / 0.05L = 60 M is the molarity I need to proceed with calculating pH and OH^-?

[Loyal]

On organic compounds when ever you write hydrogens on the inside you are telling your reader that those hydrogens are part of the hydrocarbon or other organic groups and not acids.  These hydrogens are most inert and even the ones who do reach are often weak.

[vort3x]

Ok... I don't see how that helps me? My question is:

So 3 mol of OH / 0.05L = 60 M is the molarity I need to proceed with calculating pH and OH-? I don't think that's right, since the -log of 60M to find pOH gives me a negative number.

[Loyal]

That was a response to your first post.  I didn't see your subsequent question.

For the problem

Calculate pH and [OH-] of solution:

10.0ml of 0.015M Ba(OH)2 with 40.0ml of 7.5x10-3M NaOH.

Your mol calculations seem a little off.  3 is way too much when you are dealing with mmol quantities of solution.   Go back and try that again.  Remember

mols = (molarity)*(liters of solution)

And in the case of Ba(OH)2 there are two moles of OH for every one of Ba(OH)2 dissolved. Do that for each of solutions and then add them up.  I think you will get a more accurate answer.

[vort3x]

So I found moles by multiplying the total molarity and total volume given, together. I then divided that by 0.05L... I got 9 x 10^-7 M.

Is that right? Find moles by multiplying total molarity x volume given, and divide that by 0.05L (total volume).

[Loyal]

You are getting closer, but still a few mistakes.    First do not forget that it is a 0.015 M solution of Ba(OH)2 which means that it is really a 0.030M solution of OH.   Second I think you are messing up on unit conversions because even with the error stated above I only get 9*10^-3.   

You are making progress, keep trying.   Try to type out all of your calculations so I can get a better sense of what you are doing wrong.

[vort3x]

So, 0.030 M BaOH + 7.5x10-3M NaOH = 0.0375 M

0.0375 M / 0.05 L = 0.75 M?

I'm using the initial molarities though of not just the hydroxides but also Ba and Na... so that can't be right for a final molarity of that solution if I'm just looking for OH?