Please help!! If you could provide the work as well it would be greatly apprecia

Question

Please help!! If you could provide the work as well it would be greatly appreciated. Thank you!

i safari File Edit View History Bookmarks Window Help ?? 54% CD Thu 7:37 PM east.cengagenow.com Cengage OWLv2 Online teaching and learning resource from Cengage Learning Test Ga. Buffers (part 2) 1 pt Question 2 pt 1 pt 1pt 1 pt 1 pt Question 7 @) ?pt Question 81 pt 1 pr Question 1 Use the References to access important values if needed for this question. A buffer solution contains 0.220 M acetic acid and 0.386 M sodium acctate. Question 3 Question 4 Question 5 Question 6 If 0.0453 moles of hydrochloric acid are added to 250. mL of this buffer, what is the pH of the resulting solution? (Assume that the volume change does not change upon adding hydrochloric acid) Submit Answer Try Another Version 10 item attempts remaining Progress: 59 items Due Jul 26 at 11:55 PM PreviousNext Finish Assignment Email Instructor Save and Exit 315 0

Explanation / Answer

0.220 M acetic acid means

1000 ml solution contain acetic acid = 0.220 mole

So,

250 ml solution contain acid = (0.220 × 250)/1000 = 0.055 mol

Similarly

250 ml of 0.386 M sodium acetate has

CH3COO- ion = (0.386×250)/1000 = 0.0965 mol

Now we will write the dissociation reaction for the acid

CH3COOH + H2O <----------> H3O^+1 + CH3COO-

And Ka for acetic acid is 1.8×10^-5

So, Ka =[CH3COO-][H3O^+1]/[CH3COOH]

[H3O^+1= Ka× [CH3COOH]/[CH3COO-]

Now we will use ICE table to find out the final equilibrium concentration.

*We will assume that the whole amount of acid is consumed in the reaction . So, [H3O^+1] = 0.0453 mol

[CH3COOH]. [H3O^+1]. [CH3COO-]

I. 0.055 0.0453+ from acetic acid 0.0965

C. +0.0453. -0.0453. -0.0453

E 0.1003 . x. 0.0512

Now

[H3O^+1] = x = (1.8×10^-5)×0.1003/0.0512 =3.53×10^-5

As we know pH = -log[H3O^+1]

= - log(3.53×10^-5) = 4.45

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