Acid-base equilibrium question titration

  • Thread starter Nelo
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  • #1
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Homework Statement



6.83 mL of a solution of NaOH is standardized against 3.06 g of potassium hydrogen phthlate
(KHC8O4H4). Calculate the concentration of the base.

Homework Equations



c=n/v
M of KHP = 198.201




The Attempt at a Solution



Here is what I did.
Calculated moles of pottassium hydrogen phthalate

3.06/198.201 = 0.0154 mol.
Used the volume of naoh to calculate concentration.

c=n/v . c= 0.0154/ 0.00683

c= 2.25 mol/L

The answer is 2.15. What have i done wrong?
 

Answers and Replies

  • #2
Borek
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Must be mistake in the key.

But the answer is not 2.25, it is 2.26. Don't round down intermediate values.
 
  • #3
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I have another question here.

"Calculate the pH of a buffer solution made by mixing 0.14 M KIO3 and 0.18 M HIO3"

The question is so vague, I have no idea where to start. All i know is that the ka of hio3 is 0.170 *from google*... and the ice table.



My attempt: assuming its a neutrilization i did Mhio3 - Mkio3 = 0.04 .

so hio3 remains.

Hio3 + h2o ----> -ion h3o+

0.04-x / x x


Ka being 0.170 = x^2/0.04-x

Since the 500 rule doesnt work, I did the quadratic formula.

Giving me -x2 + 6.8*10^-3 -0.170x ( -x^2 -0.170x +6.8*10^-3)

Used the formula.

calculating 0.0334 M

-log of that is 1.47. however the answer is 0.66

What have i done wrong?
 
  • #4
Borek
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Without volumes (or at least their ratio) given question can't be answered.

Regardless of whether the question can be answered or not, your ICE table is wrong. KIO3 is fully dissociated, which means initial concentration of IO3- is not zero.

Please use correct capitalization, things like hio3 or -ion don't mean anything.
 
  • #5
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Well, ... what about this one? ( Im doing exam review questions, no ka charts or any of the such are really given , theyre pretty awful. Sorry if im being annoying, I just want to know how to do all of these)

Quest. : "What volume of 0.765 M H3PO4 is required to exactly neutralize 2.000 g of calcium hydroxide?"

My attempt: 2g of caoh / 57.084 = 0.0350mol of caoh.

used stoic (balanced eq. was not given) 2h3po4 + 3caoh -----> Ca3(PO4)2 + 6h2o

2/3 * 0.0350 = 0.02333mol of h3p04.

V= n/c 0.02333/0.765 = 0.0304L. Which is wrong. The ansswer is.. : 0.0235.

Notice anything wrong?
 
  • #6
Borek
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28,777
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Notice anything wrong?

At least two things.

First, with each new post you ignore earlier comments. Yesterday I told you to not round down intermediate results, today you started with the same mistake. 10 minutes ago I told you to use correct capitalization of formulas, now you write things like caoh. If you are seriously thinking bout getting help, don't ignore it, or you will be not treated seriously.

Second, caoh is wrong not only because of incorrect capitalization.
 
  • #7
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Thanks, figured it out
 

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