If 28 mLs of a 0.25 molar NaOH solution is used to titrate 14 mLs of a nitric acid to its phenolphthalein endpoint what is the molar concentration of the acid

1. Write the balanced equation between occurring.

2. Calculate the number of moles of NaOH used for titration.

3. Calculate the number of moles of HNO3 in the 14 mL using your balanced equation.

4. From the volume of the acid and the number of moles of acid present in it, find the concentration of the acid.

Post what you get :)

well I balanced my equation NaOH+HNO3--> H2O+NaNO3
then the number of moles I got for NaOH was 40 and for HNO3 I got 63 but I don't understand how to put it together

Yes, the equation is good. Now, you obtained the wrong number of moles of both substances.

1. To get the number of moles of NaOH in 28 mL of 0.25 M NaOH, you basically use proportions.

In 1000 mL, there are 0.25 mol
In 1 mL, there would be 0.25/1000 mol
Hence, in 28 mL, there will be (0.25/1000)*28 = 0.007 mol

2. From your equation, 1 mol of NaOH will react with one mol of HNO3.
How many moles of HNO3 will react with 0.007 mol of NaOH?

3. Then, you do the inverse of the first working I showed to get the concentration of HNO3, that is:

14 mL of HNO3 contains x moles (the number of moles of HNO3 that you got from step 2)

1 mL of HNO3 will contain ? Moles of HNO3

Hence, 1000 mL of HNO3 contains ? Moles of HNO3
This value is also the concentration.

Can you work it out? Post what you get :)

1000 mL of HNO3 contains .014 moles. So I think I would have a final equation that looks like this:

.007 divided by 0.14 equals .5 M concentration. Is that correct?

Um... I don't know why you said that 1000 mL of HNO3 contains 0.014 moles... but you got the right final answer.

From your equation, 1 mol of NaOH reacts with 1 mol of HNO3

Hence, 0.007 mol of NaOH will react with 0.007 mol of HNO3.

So, you have 0.007 mol for HNO3 in 14 mL of acid.

In 1 mL, you have 0.007/14 mol
In 1000 mL, you have 0.5 mol

The concentration is 0.5 M.