Lab-04-D Graphing Acids-Bases- -p H Answers PDF

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LAB 4: GRAPHING; ACIDS, BASES & pH NAME:

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PID:

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LAB SECTION: ______________________________________ DATE:

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Points

PART A: GRAPHING, STANDARD CURVES, UNKNOWN SOLUTIONS In this exercise, we will first calculate the concentrations of four standard solutions from data provided. Next, we will calculate the densities of these standard solutions from measurements of their weights. We will then construct a standard curve graph and use it to determine the concentrations of unknown solutions whose densities were also determined. The goals of this exercise are to understand the concept and usefulness of standard curves and acquire practical experience in making and using graphs to analyze and interpret scientific data.

Table 1 Answers: 1) 0 / 1 2) 0 / 1 3) 0 / 1 4) 0 / 1

Table 2 Answers: 1) 0 / 1 2) 0 / 1 3) 0 / 1 4) 0 / 1

Table 1: Preparation of four standard solutions. Final Standard Concentration NaCl Volume Solution % NaCl (w/v) (g) (mL) # 1 0 0.0 50 2 5 2.5 50 3 10 5.0 50 4 15 7.5 50 Table 2: Density of the four standard solutions. Standar Mass of 10.0 Density d mL of (g/mL) Solution solution (g) # 0.985 1 9.85 2 10.25 1.025 3 10.58 1.058 4 10.96 1.096 1. Calculate the % concentration (w/v) of the four standard solutions in Table 1. Show your work below and put the answers in Table 1 above.

Table 1 Work 0 / 0.5 / 1

Standard 1: (0.0 g NaCl/50mL) = 0.0 g/mL x 100 = 0 % NaCl (w/v) Standard 2: (2.5 g NaCl/50mL) = 0.05 g/mL x 100 = 5 % NaCl (w/v) Standard 3: (5.0 g NaCl/50mL) = 0.1 g/mL x 100 = 10 % NaCl (w/v) Standard 4: (7.5 g NaCl/50mL) = 0.15 g/mL x 100 = 15 % NaCl (w/v)

Table 2 Work 0 / 0.5 / 1

2. To determine the densities of the four standard solutions, the mass of 10.0 mL of each was measured using a balance. Data are shown in Table 2 above. Use these data to calculate the density of each of the standard solutions. Show your calculations below and put the answers in Table 2 above. Standard 1: (9.85 g/10mL) = 0.985 g/mL Standard 2: (10.25 g/10mL) = 1.025 g/mL Standard 3: (10.58 g/10mL) = 1.058 g/mL

________pts /10 pts

Standard 4: (10.96 g/10mL) = 1.096 g/mL 2

Graph: 4 solutions 0 / 0.5 / 1 X-axis & label 0 / 0.5 / 1 Y-axis & label 0 / 0.5 / 1 Best-fit line 0 / 0.5 / 1 Line equation 0 / 0.5 / 1

3. Prepare a standard curve of concentration (% NaCl) vs density (g/mL) for the four standard solutions. You will use the data you calculated in Table 1 and Table 2, respectively. See Lab 4 document for an example of what yours should look like. Make sure that your axes are labeled (including units) and that you have a ‘best-fit’ line and the equation of that line shown on the graph. Print your graph and attach it to this report. Example of what their graph should look like

4) Work: 0 / 0.5 / 1

4. To determine the unknown concentrations of three solutions A, B, and C, the mass of 10.0 mL of each was determined (shown below in Table 3). Calculate the densities of these three unknown solutions. Show your work below and record the answers in the Table 3 below. Unknown A: (10.78 g/10mL) = 1.078 g/mL Unknown B: (9.94 g/10mL) = 0.994 g/mL Unknown C: (10.36 g/10mL) = 1.036 g/mL

5) Work: 0 / 0.5 / 1

5. You the equation of the ‘best-fit’ line from the standard curve (problem 3), as well as the density values of the unknown solutions you calculated (problem 4) to determine the % (w/v) concentrations of each of the unknown solutions. Show your calculations below and record the answers (to 1 decimal place) in Table 3 below. Equation of the line: y = 0.0073x + 0.9861 so x = (density – 0.9861)/(0.0073) Unknown A: x = (1.078 – 0.9861)/(0.0073) = 12.6 Unknown B: x = (0.994 – 0.9861)/(0.0073) = 1.1 Unknown C: x = (1.036 – 0.9861)/(0.0073) = 6.8

Table 3 Answers- Density: A) 0 / 1 B) 0 / 1 C) 0 / 1 Answers- Conc % A) 0 / 1 B) 0 / 1 C) 0 / 1

Table 3: Density & concentration of three unknown solutions. Concentration Unknow Mass of Density % (w/v) n 10.0 mL of (g/mL) Solution solution (g) 1.078 12.6 A 10.78 B 9.94 0.994 1.1 C 10.36 1.036 6.8

_______pts /13 pts 3

1a) Discussion 0/1/2

1b) Discussion 0/1/2

2a) 0 / 0.5 / 1 2b) Answer 0 / 0.5 Explanation 0 / 0.25 / 0.5 2c) Answer 0 / 0.5 Explanation 0 / 0.25 / 0.5 2d) Answer 0 / 0.5 Explanation 0 / 0.25 / 0.5

_______pts /8 pts

PART B: SPECIFIC GRAVITY & DENSITY 1. The specific gravity values of urine were determined by the 2005 BIOC107 students and are shown in the graph. Some values were determined from morning samples (first urine upon arising), whereas others were from afternoon samples. a. Briefly discuss one or two factors that might account for the large degree of variability in individual student urine specific gravity values. The wide variation is expected for any clinical parameter, but there is probably a little extra variation because of college life-styles. Some factors might be how long since last urination, how much and what they had to drink that day, etc. As long as they put thought into it, give them credit. b. Why do you think the average morning specific gravity value is somewhat higher than the average afternoon value? Morning urine is on average denser because a whole night’s waste needs to fit in one bladder’s worth of urine. During the day, there’s more fluid intake (and excretion), so on average, the waste gets more diluted. There is probably also some fluid conservation or retention during sleep so you don’t get dehydrated, another reason for higher specific gravity of overnight urine. 2. An object has a mass of 1.020 kg and dimensions of 5 cm x 10 cm x 15 cm. a. What is the density of the object? 1.020 kg x (1000 g/kg) / (5 x 10 x 15 cm3) = 1.36 g/cm3

b. Would this object float in Dr. Temple’s tropical fish tank (fresh water)? Why or why not? No, because the object is denser than fresh water 1.36 g/cm3 > 1.0 g/cm3

c. The density of seawater off North Topsail Beach, NC is 1.029 g/cm3. Would the object float in the ocean? Why or why not? No, because the object is denser than salt water 1.36 g/cm3 > 1.029 g/cm3

d. Would it float in a beaker of mercury? Why or why not? Yes, because the object is less dense than mercury 1.36 g/cm3 < 13.5 g/cm3

4

_______pts /20 pts PART C: BIOCHEMICAL CALCULATIONS ON ACIDS, BASES, and pH The best way to become familiar with the concepts of acids, bases, pH, and buffers is to work problems involving these concepts.

Table Answers (1 pt/ box) A) 0 / 1 / 2 / 3 / 4 B) 0 / 1 / 2 / 3 / 4 C) 0 / 1 / 2 / 3 / 4 D) 0 / 1 / 2 / 3 / 4

1. Complete the following table by calculating the missing entries and indicate whether the solution is acidic or basic. Show all of your work below. Be sure your answers are expressed to the proper number of significant figures. acidic or [H3O+] [OH-] pH pOH basic? 4.2 < 7 = 1.7 x 10-10 M 4.2 9.8 A 5.8x10-5 M acidic -9 8.8 > 7 = * 2.0 x 10 M or B 6.2x10-6 M 8.8 5.2 1.6 x 10-9 M basic 11.24 > 7 = C 5.75 x 10-12 M 1.74 x 10-3 M 11.24 2.76 basic 4.3 < 7 = D 5.01 x 10-5 M 2.00 x 10-10 M 4.30 9.70 acidic -8.8 * some calculators show 0.000000002 on the screen for 10 , and some show 1.6 x 10-9, so the calculator may round off without the student being aware of it. Note: There are several ways for the students to work through the calculations so answers may vary slightly. (i.e. You may round differently depending on whether you are going from a pH of 8.8 to [H+] using antilog, or from a concentration of [OH-] of 6.2x10-6 M using [H+][OH-] = 10-14. While I have shown one way of solving for the missing entries, there are other ways. As long as their work is present and makes sense, give them credit. Show your calculations here for each row here: A: [OH-] = [H+][OH-] = 10-14 = 10-14/[5.8x10-5 M] = 1.7 x 10-10 M pH = pH = -log[H+] = -log[5.8x10-5 M] = 4.2 pOH = pH + pOH = 14 = 14 - 4.2 = 9.8

Work: A) 0 / 0.5 / 1 B) 0 / 0.5 / 1 C) 0 / 0.5 / 1 D) 0 / 0.5 / 1

B: [H3O+] = [H+][OH-] = 10-14 = 10-14/[6.2x10-6 M] = 1.6 x 10-9 M pH = pH + pOH = 14 = 14 - 5.2 = 8.8 pOH = pOH = -log[OH-] = -log[6.2x10-6 M] = 5.2 C: [H3O+] = 10-pH = 10-11.24 = 5.75x10-12 M [OH-] = 10-pOH = 10-2.76 = 1.74x10-3 M pOH = pH + pOH = 14 = 14 - 11.24 = 2.76 D: 5

[H3O+] = 10 = 10-4.30 = 5.01x10-5 M [OH-] = 10pOH = 10-9.70 = 2.00x10-10 M pH = pH + pOH = 14 = 14 – 9.70 = 4.30 -pH

2) Work 0 / 0.5 / 1 Answers (1 each) 0/1/2/3

Answers (1 each) 0/1/2/3/4

_______pts /19 pts 2. A gastric juice sample from a patient suspected of having a gastric ulcer due to H. pylori has a pH value of 5.3 (normal is pH 1.0-2.0). What is the pOH, [H3O+] and [OH-] of the sample? pOH = 14 – 5.3 = 8.7 [H3O+] = 10-pH = 10-5.3 = 5.0x10-6 [OH-] = 10-pOH = 10-8.7 = 2.0x10-9 3. What is the pH of the following solutions of strong acids and strong bases (assume complete ionization, e.g., all H+ and Cl– and no HCl). Write the dissociation equations for HCl, HNO3, KOH and NaOH. a. 75 µM HCl

3) Work 0 / 0.5 / 1 Answers (1 each) 0/1/2/3/4 Equations (1 each) 0/1/2/3/4

pH = -log[H+] pH = -log(75x10-6 M) = 4.12

HCl  H+ + Clb. 5 x 10–4 M HNO3 pH = -log[H+] pH = -log(5x10-4 M) = 3.3 HNO3  H+ + NO3c. 0.2 M KOH

pOH = -log[OH-] pOH = -log[0.2 M] = 0.70 pH = 14 – pOH = 14 – 0.70 = 13.3

KOH  K+ + OHd. 4.56 x 10–3 M NaOH pOH = -log[OH-] pOH = -log[4.56x10-3 M] = 2.34 pH = 14 – pOH = 14 – 2.34 = 11.66 NaOH  Na+ + OH-

4) Answer 0 / 0.5 Explanation 0 / 0.25 / 0.5

4. Does pH of a solution change with temperature? Why? (you can use the internet to help you answer this question) Yes, the pH changes with temperature as the molecules have higher energy and are more easily ionized, thereby increasing the [H+] concentration for an acid or the [OH-] concentration for a base. For an acid, [H+] will increase and pH decrease. For a base, [OH-] will increase and pH increase. 5. At 30°C, the value of Kw (water constant) is 1.5x10-14. Determine the pH, pOH, [H+], and [OH-] for water at 30°C.

5) Work 0/ 0.5 / 1

[H+][OH-] = 1.5 x 10-14 AND [H+] = [OH-] because pure H2O dissociates to give 1 H+ for every 1 OH6

Two ways to calculate it: (1)

[H +] = [O H-] = sq ua re ro ot (1. 5x 1014 ) = 1. 22 x1 0-7 M an d pH = pO H = -lo g( 1. 22 x 107 ) = 6. 91

(2)

pH = pOH = (-log(1.5x10-14))/2.0 = 6.91 and [H+] = [OH-] = 10(-6.91) = 1.23 x 10-7 M

6) Work 0/ 0.5 / 1 Answers (1 each) 0/1/2/3/4

7) Work 0/ 0.5 / 1 Answers (1 each) 0/1/2/3/4

8) Work 0/ 0.5 / 1 Answers (1 each) 0/1/2/3/4

Intro 0 / 5 / 10 / 15

OR

7

_______pts /30 pts

6. Normally, rain has a pH of approximately 5.5. a. What is the [H30+] of rain with a pH=5.5 ? [H3O+] = 10 = 10(-5.5) = 3.2 x 10-6 M

pOH = 14 – pH = 14 – 3.2 = 10.8 [OH-] = 10-pOH = 10-10.8 = 1.6 x 10-11 M 7. 25 ml of 6 M NaOH was diluted with water to 400 ml. What is the pH and pOH of the final solution? What is the [H+] and [OH-]? (Remain calm; remember the formula for dilution from the previous lab session: (V1)(C1) = (V2)(C2) ) (25 mL)(6 M) = (400 mL) X = 0.375 M NaOH (final solution concentration) [OH-] = 0.375 M pOH = -log(0.375) = 0.43 pH = 14-0.43 = 13.57 [H+] = 10-13.6 = 2.7 x 10-14 M 8. A solution was made by dissolving 10 g of NaOH (formula weight 40 g/mole) in water and the final volume brought to 500 ml. What is the pH, pOH, [H+] and [OH–] of this solution?

-pH

b. A study about acid rain in Scranto n, PA determi ned the average pH value of acid rain in this area to be 3.2. What is the average H30+ and OHconcent ration of this acid rain?

10 g NaOH / (40 g/mole) / (0.5 L) = 0.5 M NaOH [OH-] = 0.5 M pOH = -log(0.5) = 0.30 pH = 14 – pOH = 13.70 [H+] = 10-13.70 = 2.0 x 10-14 M

INTRODUCTION PARAGRAPH Your Introduction should consist of three short paragraphs of a few sentences each. The first paragraph should briefly discuss the important of graphing data in science. The second paragraph should briefly discuss the significance of specific gravity values in clinical diagnosis and management. The third paragraph should briefly discuss acids, bases and pH. Your introduction must be typed and attached to this worksheet. Page 2 Page 3 Page 4 Page 5 Page 6 Page 7

________ pts / 10 pts ________ pts / 13 pts ________ pts / 8 pts ________ pts / 20 pts ________ pts / 19 pts ________ pts / 30 pts

TOTAL ________ pts / 100 pts

[H3O+] = 10-pH = 10-3.2 = 6.3 x 10-4 M 8...