Your goal is to make a buffer with a pH of 4.95 from acetic acid and sodium acetate. ASsume the pKa for acetic acid is 4.74. What is the volume of acetic acid added make your buffer? Report your answer to the tenths place. To do this, we use the Henderson-Hasselbach Equation pH = pkA + log (base/acid) where pkA = 4.74 for acetic acid, pH = target pH and [base] /[acid] = x/(1-x) since we don't know the concentration of the base, we will call it x the acid is just 1-x, assuming the two add up to 100% solve for x, and then you know the ml of base and acid to add.

Your goal is to make a buffer with a pH of 4.95 from acetic acid and sodium acetate. ASsume the pKa for acetic acid is 4.74. What is the volume of acetic acid added make your buffer? Report your answer to the tenths place. To do this, we use the Henderson-Hasselbach Equation pH = pkA + log (base/acid) where pkA = 4.74 for acetic acid, pH = target pH and [base] /[acid] = x/(1-x) since we don't know the concentration of the base, we will call it x the acid is just 1-x, assuming the two add up to 100% solve for x, and then you know the ml of base and acid to add.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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2.74 g of canola oil were titrated with KOH. It required 13.92 mL of 0.0040 M KOH to reach the equivalence point. What is the acid number? Your Answer: Answer units
1.What is the Ka of a weak acid, HA, that forms a buffer of pH 5.50 when combined in an equal concentration with its conjugate base, A−? Answer: Ka= ...... x 10^-6 2.Calculate the pH of 0.136 M potassium nitrite, KNO2. The ionization constant of nitrous acid, HNO2, is Ka= 5.6× 10-4 3.Calculate the pH of a solution of magnesium hydroxide, Mg(OH)2 (Ksp = 5.6 × 10−12), made by dissolving solid Mg(OH)2 in pure water until the solution is saturated.
We are titrating 25.0 mL 0.100 molar weak acid HA with 0.100 molar NaOH. We find that at the equivalence point, when we have added 25.0 mL of 0.100 molar NaOH, the pH is 9.5. We want to find the pka. The first step is to fill in this ICE chart. What do we put in for x in this chart? Hint, find pOH (14- pH) Then, [OH-] is 10^-pOH [OH-] is x (You do NOT have to do it, but would you know how to get pka now?? Hint, you would need to find pKb) C E A- (+ H2O) 0.050 M © 2.5x106 O 1.5 x 10-3 ○ 2.9 x 10-4 3.1 x 10-5 0.012 0.050 M-X 0 X HA OH- 0 X
4) a) Determine the mass (in grams) of sodium formate that must be added to 1.00 L of 0.180 M formic acid to prepare a buffer with a pH = 3.50. Assume there is no change in volume upon addition of the sodium formate. b) Write the predominant chemical reaction that occurs when 0.050 mol of HClO4 is added to 1.00 L of the buffer prepared in part a and calculate the pH of the solution.
8. You are titrating 50.0 mL of 0.200 M propanoic acid with 0.100 M NaOH. Use the data in the table to answer the questions. Name Propanoic acid Structure CH3CH2CO2H pka 4.874 Ka 1.34 × 10-5 a) Assuming that the equivalence point is considered 100% completion, calculate the pH at the following points: 0%, 10%, 50%, 90%, 100%, and 120%. b) Sketch the titration curve. c) What would be an ideal pKa for an indicator to use in this titration?
ou need to produce a buffer solution that has a pH of 5.03. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74. Express your answer in millimoles.
3 NEXT > < PREV 1 2 Based on your ICE table (Part 1) and the definition of Ka, set up the expression for Ka in order to determine the unknown. Each reaction participant must be represented by one tile. Do not combine terms. Ka = = 6.3 × 10-5 RESET [0] [21.5] [37.7] [0.176] [0.261] [0.880] [1.31] [0.178] [0.888] [x] [2x] [21.5-x] [37.7+x] [0.176-x] [0.261 + x] [0.880 - x] [1.31 + x] [0.178 + x] [0.888 + x] < PREV 1 2 3 Based on your ICE table (Part 1) and the equilibrium expression for Ka (Part 2), determine the pH of the buffer solution. pH = RESET 0 4.2 x 10-5 6.23 4.37 2.4 × 10-11 7.77 9.63 0.880
You have 100 mL of an acetic acid/acetate buffer. The solution is comprised of 0.33 M acetic acid and 0.01 M acetate. What is the new pH when 5.0 mL of 0.10 M HCl is added to this solution? The pKa of this pair is 4.76. (Note: You must use the I.C.E method for this one).
65 mL of 0.65 M benzoic acid (Ka = 6.4×10-5) is titrated by 0.05 M NaOH. 1)Calculate the pH of the acid solution before any titrant is added. 2)Calculate the pH after 795.9 mL of 0.05 M NaOH is added to 65 mL of 0.65 M benzoic acid. 3)Calculate the pH after 845 mL of 0.05 M NaOH is added to 65 mL of 0.65 M benzoic acid. 4)Calculate the pH after 1.58 x 10^3 mL of 0.05 M NaOH is added to 65 mL of 0.65 M benzoic acid.
You are asked to make a buffer of pH 3.4 and given the above acids. Of these acids, which would create a buffer with the best capacity against acids? Which would have the best buffering capacity against bases? Which could not be used at all for this project? Best against acids Best against bases Acid #1 has a pka of 3.15 Acid #2 has a pKa of 3.65 Acid #3 has a pKa of 4.23 Acid #4 has a pka of 4.79 Cannot be used Acid #2 Acid #1 Acid #4
where did the 3.18 come from?
In order to determine the Ka of a weak, monoprotic acid, a student measured the pH of a solution of this weak, monoprotic acid to be 3.617. Based on the full- neutralization titration portion of the experiment, it was found that the total molar concentration of this acid solution was 2.632 M. Using these measurements, what would be the Ka for this weak, monoprotic acid? Ka X 10-7 Please enter as your answer only the underlined numerical part of the Ka above. Round your answer to two decimal places.