Quiz 4 - Totalscore: 110/120=91.6667%...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Total score: 110/120 = 91.6667% Total score adjusted by 0.0  Maximum possible score: 120  1. You want to make an acetic acid/acetate buffer, pH 5.6, with a volume of 250 mL, and a final concentration of ([AcOH] +  [AcO - ]) = 0.1 M.  You may only use acetic acid and sodium acetate (no strong acid or base).  How many mL of glacial  acetic acid would be needed (d = 1.05 g/mL)?  Student  Response Value Correct Answer Feedback 1.   2.16 mL 2.   1.26 mL 3.   0.17 mL 100%  4.   0.20 mL General Feedback: Using the Henderson-Hasselbalch equation: log([A - ]/[HA]) = pH - p K a [A - ]/[HA] = 10 (pH - p K a) Our other constraint is: [HA] + [A - ] = C (in this case, 0.1M) or [A - ] = C -  [HA] Putting the two equations together: (C - [HA])/[HA] = 10 (pH - p K a) C/[HA] - 1 = 10 (pH - p K a) C/[HA] = 10 (pH - p K a)  + 1 [HA] = C/(10 (pH - p K a)  + 1) Inserting the values for this calculation: [HA] = 0.1 M /(10 (5.6 - 4.74)  + 1) [HA] = 0.012 M At this point, you can use acetic acid's molecular weight, the volume of the buffer solution,  and the density of glacial acetic acid to calculate the volume needed. Score: 10/10 
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Comments: 2. In a titration of 100 mM lactic acid (p K a  = 3.86) with 100 mM NaOH, what will the pH be at the equivalence point?  Student  Response Value Correct Answer Feedback 1.   9.4 2.   8.3 100%  3.   3.6 4.   5.7 General Feedback: See point  d  at the bottom of p. 728 in the text for how to calculate pH at the equivalence point for a  weak acid/strong base titration. Score: 10/10  Comments: 3. The range of blood pH that is considered normal in humans is 7.35 to 7.45. If a person's blood pH changes from 7.35 to  7.45, what is the change in [H 3 O + ]?  Student  Response Value Correct Answer Feedback 1.   -2.3 × 10 -8  M
Background image of page 2
Student  Response Value Correct Answer Feedback 2.   -1.3 × 10 -8  M 3.   1.3 × 10 -8  M 4.   -9.2 × 10 -9  M 100%  5.   9.2 × 10 -9  M General Feedback: The change in [H 3 O + ] = 10 -pH(final)  - 10 -pH(initial) Therefore, Δ[H 3 O + ] = 10 -7.45  - 10 -7.35 Δ[H 3 O + ] = -9.2 × 10 -9  M Score: 10/10  Comments: 4. What is the pH of the solution that results from mixing the following four solutions together?
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/13/2011 for the course CHEM 1AA3 taught by Professor H during the Spring '08 term at McMaster University.

Page1 / 11

Quiz 4 - Totalscore: 110/120=91.6667%...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online