Use the H-H equation and Table 2.6 to do the following calculations. For the solutions with...
- Use the H-H equation and Table 2.6 to do the following calculations. For the solutions with the following pH values, calculate the ratio of the concentrations of the base and the acid forms of the substances listed.
-
- for an acetic acid/acetate solution with a pH of 2.66: [A-]/[HA] = ______________
- for a phenol/phenolate solution with a pH of 10.24: [A-]/[HA] = _______________
- for a phosphate buffered solution with a pH of 6.55: [A-]/[HA] = _______________
Homework Answers
According to H-H equation:
-pKa = -pH + log([A-]/[HA])
a. pH = 2.66, pKa of acetate solution: 4.76
[A-]/[HA] = 10^(-4,76+2.66) = 0.0079
b. pH = 10.24, pKa of phenol: 10
[A-]/[HA] = 10^(-10+10.24) = 1.74
c. pH = 6.55, pKa of phosphate buffered solution: 6.85
[A-]/[HA] = 10^(-6.85+6.55) = 0.501
Add Answer to:
Use the H-H equation and Table 2.6 to do the following
calculations. For the solutions with...
21. Use the H-H equation and Table 2.6 to do the following calculations. For the solutions...
21. Use the H-H equation and Table 2.6 to do the following calculations. For the solutions with the following ratios of [A-]/[HA], calculate the pH of the solutions. a. for a solution containing 0.25M sodium phosphate monobasic and 0.87M sodium phosphate dibasic: pH = ___________________ b. for a solution containing 0.33M sodium bicarbonate (NaHCO3) and 0.55M sodium carbonate (Na2CO3): pH = ______________________ c. for a solution containing 0.55mM lactic acid and 0.88mM sodium lactate: pH = __________________________
PH of Buffered solutions: Acids, If the calculations do not fit into the table, then number the c...
pH of Buffered solutions: Acids, If the calculations do not fit into the table, then number the calculation and use a separate pH Calculations t n Sey Ayat Buffer 2 4.74 on Buffer 1A and но t . 52|1513 |0431시 A2 6 Se pava Da pv 119 Chem 24B pH of salts and buffers Revised Sp 17 Solution pH Buffer 2A and HCI 119 ow Buffer 1B and on Se vavante Buffer 2B and NaOH 5-3千 ơn pu Pev 2...
What concentrations of acetic acid (pKa=4.76)(pKa=4.76) and acetate would be required to prepare a 0.10 M0.10...
What concentrations of acetic acid (pKa=4.76)(pKa=4.76) and acetate would be required to prepare a 0.10 M0.10 M buffer solution at pH 4.6pH 4.6? Note that the concentration, pH value, or both may differ from that in the first question. Strategy Rearrange the Henderson–Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A−]/[HA][A−]/[HA] . Use the mole fraction of acetate to calculate the concentration of acetate. Calculate the concentration of acetic acid. Step 1: The ratio...
Question: In the space below, discuss tour observations and compare the use of a buffered versus...
Question: In the space below, discuss tour observations and
compare the use of a buffered versus a non-buffered solution (DI
water) with the addition of a strong acid or base.
Part C: K, Acid Dissociation Constant Acetic Acid pH | [H] | K Calculation Concentration (2.83 x 10 2.82 104 7.95 x 10-6 0.010 M 3.55 0.10 M (8.71 * 10-4) (0.1) 7.59x10-6 3.00 18.71x10 *M 1.0 M 2.71 (1.95 x 10-5) (1.03 1.95 x 10-3M 3.80 x 10-6 LHO...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.10 M buffer solution at pH 4.9? Note that the concentration and/or pH value may differ from that in the first question. 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A^-]/[HA]. 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. That is, there are...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.20 M buffer solution at pH 5.0? Note that the concentration, pH value, or both may differ from that in the first question. Strategy 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid). [A-V[HA). 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. Step 1:...
Resources Ex Give Up? What concentrations of acetic acid (pka = 4.76) and acetate would be...
Resources Ex Give Up? What concentrations of acetic acid (pka = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 4.92 Note that the concentration, the pH, or both values may differ from that in the first question. Strategy 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [ A HA). 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate...
Use the Henderson-Hasselbalch equation to perform the following calculations. The K a of acetic acid is...
Use the Henderson-Hasselbalch equation to perform the following calculations. The K a of acetic acid is 1.8 10 –5 . a. Buffer A: Calculate the mass of solid sodium acetate required to mix with 100.0 mL of 0.5 M acetic acid to prepare a pH 4 buffer. Record the mass in your data table. b. Buffer B: Calculate the mass of solid sodium acetate required to mix with 100.0 mL of 1.0 M acetic acid to prepare a pH...
The Henderson-Hasselbalch equation connects pH to pk, by relating pH to the relative amounts of the...
The Henderson-Hasselbalch equation connects pH to pk, by relating pH to the relative amounts of the acid and conjugate base. The equation is: [A], pH = pKa + log [HA]' A. If you had an acetic acid solution at pH 4.75, what would the ratio of acetic acid to acetate 4. be? (Сн,соо у сн, соон) - ([CH3CO0¯], [CH3COOH], B. What if the solution pH was 4.27? C. What about pH 5.05?
How do you get the 1.18 when doing the quadratic equation? Therefore, we have pH =...
How do you get the 1.18 when doing the quadratic
equation?
Therefore, we have pH = 4.74 + log(1.00 times 10^-4)/(1.00 times 10^-3) pH = 4.74 - 1.00 = 3.74 Therefore Now we will redo the calculation, without making any assumptions at all. We will solve for x, which represents the H^+ concentration at equilibrium, in order to calculate pH. [CH_3COO^-][H^+]/[CH_3COOH] = K_a (1.00 times 10^-4 + X)(X)/(1.00 times 10^-3 - X) = 1.8 times 10^-5 1.00 times 10^-4 X...
pH of Buffered solutions: Acids, If the calculations do not fit into the table, then number the calculation and use a separate pH Calculations t n Sey Ayat Buffer 2 4.74 on Buffer 1A and но t . 52|1513 |0431시 A2 6 Se pava Da pv 119 Chem 24B pH of salts and buffers Revised Sp 17 Solution pH Buffer 2A and HCI 119 ow Buffer 1B and on Se vavante Buffer 2B and NaOH 5-3千 ơn pu Pev 2...
Question: In the space below, discuss tour observations and
compare the use of a buffered versus a non-buffered solution (DI
water) with the addition of a strong acid or base.
Part C: K, Acid Dissociation Constant Acetic Acid pH | [H] | K Calculation Concentration (2.83 x 10 2.82 104 7.95 x 10-6 0.010 M 3.55 0.10 M (8.71 * 10-4) (0.1) 7.59x10-6 3.00 18.71x10 *M 1.0 M 2.71 (1.95 x 10-5) (1.03 1.95 x 10-3M 3.80 x 10-6 LHO...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.10 M buffer solution at pH 4.9? Note that the concentration and/or pH value may differ from that in the first question. 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A^-]/[HA]. 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. That is, there are...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.20 M buffer solution at pH 5.0? Note that the concentration, pH value, or both may differ from that in the first question. Strategy 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid). [A-V[HA). 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. Step 1:...
Resources Ex Give Up? What concentrations of acetic acid (pka = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 4.92 Note that the concentration, the pH, or both values may differ from that in the first question. Strategy 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [ A HA). 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate...
The Henderson-Hasselbalch equation connects pH to pk, by relating pH to the relative amounts of the acid and conjugate base. The equation is: [A], pH = pKa + log [HA]' A. If you had an acetic acid solution at pH 4.75, what would the ratio of acetic acid to acetate 4. be? (Сн,соо у сн, соон) - ([CH3CO0¯], [CH3COOH], B. What if the solution pH was 4.27? C. What about pH 5.05?
How do you get the 1.18 when doing the quadratic
equation?
Therefore, we have pH = 4.74 + log(1.00 times 10^-4)/(1.00 times 10^-3) pH = 4.74 - 1.00 = 3.74 Therefore Now we will redo the calculation, without making any assumptions at all. We will solve for x, which represents the H^+ concentration at equilibrium, in order to calculate pH. [CH_3COO^-][H^+]/[CH_3COOH] = K_a (1.00 times 10^-4 + X)(X)/(1.00 times 10^-3 - X) = 1.8 times 10^-5 1.00 times 10^-4 X...
Most questions answered within 3 hours.
-
Where is the error in this code sequence?
String s1 = "Hello";
String s2 = "ello";...
asked 10 months ago -
Financial data for Joel de Paris, Inc., for last year
follow:
Joel de Paris, Inc.
Balance...
asked 10 months ago -
Consider this reaction:
Al2(SO4)3 (aq)+ BaCl3
(aq) Al2Cl6 (aq)- +
3BaSO4(s) . What is the...
asked 10 months ago -
Suppose that Savneet is considering increasing her
recent random sample from 20 car rentals to 40...
asked 10 months ago -
Trucks arrive at an unloading terminal at an average rate of 120
per hour.
Trucks arrive...
asked 10 months ago -
Why are methanol and ethanol completely soluble in water while
octanol is not very little soluble....
asked 10 months ago -
A facilities manager at a university reads in a research report
that the mean amount of...
asked 10 months ago -
When the CuSO4 is rehydrated by adding water to the anhydrous
compound, is this an endothermic...
asked 10 months ago -
A ray of sunlight is passing from diamond into crown glass; the
angle of incidence is...
asked 10 months ago -
A block of mass 0.249 kg is placed on top of a light, vertical
spring of...
asked 10 months ago -
how do the kidneys compensate in the presences of acidosis
a) trigger hyperventilate
b) reserve acid...
asked 10 months ago -
Question 501 pts
The rental rate of capital to the firm increases. Which of the
following...
asked 10 months ago