Yes to the ICE table (at least a modified one) and no, there is no limiting reagent.
First you want to determine the moles HCl and moles H3PO4 mixed. M x L = moles.
All of that is added to a 1 L volumetric flask, diluted to the mark. So what is the concn of HCl and H3PO4. Obviously, its just moles HCl/1000 and moles H3PO4/1000. Now you take 50 mL of that, so M x L again gives you moles of the titrated amount in the 50 mL portion. That's the hard part. The easy part is next.
Remember H3PO4 is a weak acid with three H ions to titrate but HCl is a strong acid. So the HCl will be titrated first, then the first H of the H3PO4, then an equal amount for the second H of the H3PO4. Add amount HCl + H #1 H3PO4 + H #2 for H3PO4 and that is the total volume for th HCl plus two hydrogens from the H3PO4. Technically, the third H from H3PO4 can't be titrated; however, it can be if one adds Ca^+2 [to ppt all of the phosphate ion as Ca3(PO4)2]
Check my thinking. Post your work if you get stuck.
Question: A volume of 26.00ml of 4.30 M. HCL is mixed with 33.70 ml of 10.80 M H3PO4 s mixture is diluted to 1.00 liter with distilled water. A 50.00 ml portion of this diluted mixture is than titrated with .2200 M. NaOH. a) compute the concentration of phosphoric acid after dilution to 1.00 liter. B) During the titration how many ml. of the NaOH will be needed to reach the second equivalence point? (all HCL titarted plus two protons from H3PO4)
I'm not really sure how i would go about solving both parts a and b. Do you need to set up and ice table? Do you have to find the limiting reactent?
3 answers
So far i got .118 mol of HCl than divided by 1000 which got me 1.18e-4 than i times that by .050 L =5.59E-6 moles. i than did the same thing for the H3PO4 and got for the final answer 1.82E-5. I'm confused on your meaning for than getting the concentration of the two. Right now they are both in moles i need to get them into molarity and than plug into ice table right. i than need the ka value right? or is this not a correct way to do it?
We're not on the same page.
If you have 0.118 moles HCl in 1 L, then the molarity is #mols/L so it must be 0.118 M. (After re-reading my first response, I probably mis-led you. I was calculating mols/mL and when multiplied by the 50 mL aliquot you would have the moes in the final mixture). The concn H3PO4 is # moles/L = 0.03370 x 10.80 M = 0.3640 moles/L = 0.3640 M.
If we take 50 mL of that 1 L solution, we now have M x L = 0.118 M x 0.050 L = ?? moles HCl and we have 0.3640 M x 0.050 L = ?? moles H3PO4.
Now you titrate that mixture with 0.2200 M NaOH. Neutralize the HCl first, then the first H on the H3PO4, then the second H on the H3PO4 and add the 3 volumes together.
If you have 0.118 moles HCl in 1 L, then the molarity is #mols/L so it must be 0.118 M. (After re-reading my first response, I probably mis-led you. I was calculating mols/mL and when multiplied by the 50 mL aliquot you would have the moes in the final mixture). The concn H3PO4 is # moles/L = 0.03370 x 10.80 M = 0.3640 moles/L = 0.3640 M.
If we take 50 mL of that 1 L solution, we now have M x L = 0.118 M x 0.050 L = ?? moles HCl and we have 0.3640 M x 0.050 L = ?? moles H3PO4.
Now you titrate that mixture with 0.2200 M NaOH. Neutralize the HCl first, then the first H on the H3PO4, then the second H on the H3PO4 and add the 3 volumes together.