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# buffer solution challenged by HCl

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I have four problems that I don't understand how to figure out. I've done problems 16-20 but don't know how to do 21-24.
I've attached the whole page from the exam for the information it provides. I've boxed my previous answers in case they are needed to figure out the four questions.
I need help on the third page. The second page is a reference to the first part of the problem.

Given 100 mL of the H2C6H8O6/NaHC6H8O6 buffer you just made, answer the following questions about the effect of a challenge to your buffer when 1.0 mL of 1.2 M HCl (stomach acid) is added to that buffer.

21. After the challenge with HCl, the pH of your buffered solutoin is
22. After the challenge with HCl, the moles of HCl remaining in your buffer is
23. After the challenge with HCl, the acidity of your buffered solution is
24. After the challenge with HCl, the final H+ concentration of your buffered solution is

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The facts: Molarity of C6H8O6 is 0.02 M
Molarity of NaC6H7O6 is 0.04 M

So in 100 ml buffer solution, the moles of C6H8O6 is (100 ml)(0.02 M) = 2 mmol
The moles of NaC6H7O6 is (100 ml)(0.04 M) = 4 mmol

The moles of HCl added is (1 ml)(1.2 M) = 1.2 mmol
We can ignore the change in the volume of the solution, that is, the final volume is still 100 ml after HCl is added.

The reaction of NaC6H7O6¬ with HCl
NaC6H7O6 + HCl NaCl + C6H8O6
Initial 4 1.2 2
Change -1.2 -1.2 +1.2 +1.2
Final 2.8 0 1.2 3.2
HCl is only 1.2 mmol, so it is the limiting reactant. All HCl reacts with NaC6H7O6 to produce C6H8O6.

After HCl is added, the moles of NaC6H7O6 is 4 - 1.2 = 2.8 mmol
The moles of C6H8O6 is 2 + 1.2 = 3.2 mmol

The molarity of NaC6H7O6 is 2.8 mmol / 100 ml = 0.028 M
The molarity of C6H8O6 is 3.2 mmol / 100 ml = 0.012 M

So the answer to 19 is 3.2 mmol or 0.0032 mol. Answer is A.
Answer to 20 is 0.0028 mol. Answer is D

21. It is the same as No. 14.
The pH of the buffer solution is

22. E

23. The H+ concentration after HCl challenge is
The original H+ concentration in the buffer is (pH = 4.4, see No. 14)
So the acidity increases by