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Please calculate the pH of a 0.5L solution prepared with 0.05 M acetic acid (pK = 4.76) before and after the addition of 2 mL of 5 M NaOH. In a secondary calculation, please calculate the pH of this solution if the buffer was not present.
Hint: 1 Do not forget to calculate the equilibrium concentrations of acid and conjugate base as a first step.
Hint: 2 Your goal is to calculate 3 pH values
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- Your goal is to make a buffer with a pH of 4.75 from acetic acid and sodium acetate. Assume the pKa for acetic acid is 4.74. What is the volume of 1 M sodium acetate added to 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 to 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.Construct a curve for the titration of 50.00 mL of a 0.1000 M solution of compound A with a 0.2000 M solution of compound B in the following table. For each titration, calculate the pH after the addition of 0.00, 12.50, 20.00, 24.00, 25.00, 26.00, 37.50, 45.00, 49.00, 50.00, 51.00, and 60.00 mL of compound B.where, Compound A (H2SO3) pka1 = 1.23 x 10^-2 pka2 = 6.6 x 10 ^-8 Compound B (NaOH
- In this lab, you will make a buffer out of acetic acid, H3CCOOH, and acetate, H3CCOO–. By convention, weak acids are often shortened to HA and their conjugate weak bases are shortened to A–. You will make a pH 5.00 buffer. Use the Henderson-Hasselbach equation to calculate the ratio of conjugate base to acid ([A–]/[HA}) in this buffer. The pKa of acetic acid is 4.75. Enter your answer with three digits.Out of the potential unknowns in this experiment, which one would be the best choice to create a buffer with a pH of 7.0? Note: Copying and/or posting this question without the express written permission from Dr. Burke and the Department of Chemistry & Biochemistry at the University of Delaware is a violation of intellectual property copyright law. Out of the potential unknowns in this experiment, which one would be the best choice to create a buffer with a pH of 7.0? Formic acid Sodium bisulfate Phenol Hypobromous acid Boric acid Sodium bicarbonate Potassium hydrogen phthalate Sodium bisulfite Disodium hydrogen citrate Ascorbic acidA. What two compounds would be best to use in preparing a pH 9.0 buffer solution? justify your answer. Assume you have the acids and the sodium salts of their conjugate bases available to you. Answer: The two compounds best fo preparing a buffer solution with a pH of 9.0 would be hydrocyanic (HCN) and phenol (HOc6H5) because their pKa values are 9.40 and 9.80 respectively making their pH levels the closest to 9. B. WHat mass ration of sodium to acid is needed to prepare this buffer? C. Can this buffer solution better resist pH changes from a string acid addition or strong base addition?
- Does my math add up? Calculate the moles of NaOH delivered at the equivalence point as determined by the second derivative plot for your careful trial, then enter the values in Table 2. 24.178(mL) x 1 / 1000 (mL) x 0.1002(M) = 0.0024709916 mol = 2.47 x 10 ^ -3 mol Calculate the number of moles of HCl as experimentally determined by the titration into Table 2. 2.47 x 10 ^ -3 mol Calculate the concentration of the original HCl solution for then enter the values in Table 2. M = n / VL = 2.47 x 10 ^-3 / 10 x 10 ^-3 = 0.247 molar Table 2. Second Derivative Processed Data Trial 2 moles of NaOH at equivalence point 2.47 x 10 ^ -3 moles of HCl 2.47 x 10 ^ -3 concentration of original HCl solution (M) 0.24725.00 mL of a saturated Ca(OH)2 sample requires 22.50 mL of 0.0250 M HCl to neutralize it. Calculate the value for Ksp of Ca(OH)2 from this data. Given the equilibrium equation: Ca(OH)2 (s) ↔ Ca2+ (aq) + 2OH– (aq) 1) Write the Ksp expression? 2) Use titration data to determine moles of OH¯ in the 25.0 mL sample (Remember, every one H+ neutralizes one OH¯.) 3) Use moles of OH¯ and sample volume to determine [OH¯]: 4) Determine [Ca2+]: 5) Calculate the Ksp for Ca(OH)2Use the titration curve for the weak acid to calculate the pH of a 0.150 M solution of that weak acid.Round the pKa to the nearest whole number for the calculation, and enter your answer for the pH to two significant figures.
- Bonus (2 points) Buffer solutions resist dramatic changes in pH due to the presence of a weak acid-conjugate base pair present in the solution. However, the combination of 20.0mL of 0.30M HC2H3O2 and 10.0mL of 0.30M NaOH also produces an excellent buffer solution. How? Provide a thorough explanation that explains the apparent discrepancy. (Using calculations and reactions to support your explanation is encouraged. Feel free to use another page if additional space in needed.) The Ka of HC2H3O2 is 1.8x 10-5.Q1.0 When 0.39 mol of solid CO2 is reacted with 0.23 mol of CH3MgBr in an ether solvent, the resulting product is isolated in aqueous solution (assuming 100% yield) and any excess CO2 is removed. Following this, 0.13 mol of HCl (aq) is added. What is the resulting solution's pH? Please note that you will need to research a pKa value for this query, and your answer should be reported to two decimal points.A 1.0L buffer is made at 25°C that consists of 0.80 mol of a weak acid HA and 0.80 mol of its conjugate base A-. The acid ionizarion constant is Ka= 1×10^-4. To this buffer solution 0.80 mol NaOH is added. Assuming that the volume change due to the added NaOH is negligible, calculate pOH in the resulting solution.
