Hi,may I have your help?

I have posted a titration question that I'm trying to solve. The correct answer to the problem is 0.128 M, however,I have arrived at the answer: 0.000128 M. Here's the question:

If 38.30 mL of 0.250 M NaOH is used to titrate 25.0 mL of phosphoric acid,what is the molarity of the acid? The balanced equation is

H3PO4(aq)+3 NaOH(aq)--- Na3PO4(aq)+3H2O (l).

Here's how I've tried to solve for the answer:
38.3 mL solution * (0.250 mol NaOH/1000 mL)(1 mol H3PO4/3 mol NaOH)= 3.19 moles H3PO4

(3.19 mol H3PO4/25.0 mL solution)(1L\1000 mL)= 1.276*10^-4
=0.000128 M

I appreciate your help.

0.0128 M is correct.

Thanks for showing your work.
Your error is in the first line. Instead of 3.19 moles, you never divided by the 1000 mL in the second term, and it should be 0.00319 moles. Then 0.00319 moles/0.025L = 0.12767 = 0.128 M.

I'd be happy to help you with the titration question!

Let's go through the calculation step-by-step to determine the correct molarity of the acid.

First, let's calculate the number of moles of NaOH used in the titration:

38.30 mL * (0.250 mol NaOH / 1000 mL) = 0.009575 mol NaOH

Since the balanced equation shows a 1:1 molar ratio between NaOH and H3PO4, we know that the number of moles of H3PO4 is also 0.009575 mol.

Next, let's convert the volume of the phosphoric acid solution to liters:

25.0 mL = 25.0 mL * (1 L / 1000 mL) = 0.0250 L

Now, we can calculate the molarity of the acid (H3PO4):

Molarity = moles of solute / volume of solution in liters
Molarity = 0.009575 mol / 0.0250 L = 0.383 M

Therefore, the correct molarity of the acid is 0.383 M.

It seems that you made an error in converting the calculated moles of H3PO4 to the final molarity. Instead of dividing by 25.0 mL, it should be divided by 0.0250 L to get the correct answer.

Sure, I'd be happy to help you with this titration problem.

To find the molarity of the phosphoric acid (H3PO4), you need to use the balanced equation and the stoichiometry of the reaction.

First, let's calculate the number of moles of NaOH used in the titration:

38.30 mL of 0.250 M NaOH = 0.0383 L * 0.250 mol/L = 0.009575 mol NaOH

According to the balanced equation, the ratio between NaOH and H3PO4 is 3:1. Therefore, 0.009575 mol NaOH will react with one-third of that amount of H3PO4.

0.009575 mol NaOH * (1 mol H3PO4 / 3 mol NaOH) = 0.003192 mol H3PO4

Now, calculate the molarity of H3PO4:

Molarity = moles of solute / volume of solution in liters

Molarity = 0.003192 mol H3PO4 / 0.0250 L solution = 0.12768 M

So, the molarity of the phosphoric acid is approximately 0.128 M, which matches the correct answer.

It seems like there was a calculation error in the steps you provided. Instead of dividing by 25.0 mL, you should have divided by 0.0250 L (which is equivalent to 25.0 mL). This resulted in the incorrect answer of 0.000128 M.

I hope this clarifies the issue and helps you understand how to solve the problem correctly. Let me know if you have any further questions!