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16. Select the true statement below regarding an A?/HAbuffer.

(1) thepurpose of the buffer is to keep [HA] and [A?] asconstant as possible

(2) HAmust be a stronger acid than H3O+ andA? must be a stronger base thanOH?

(3) strong base added to an effectivebuffer solution will be consumed by HA

(4) the optimal buffer will have low[HA] and[A?] (5)the pKa of HA must beequal to pH of the buffer


Please explain why each is wrong and which is correct.

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Bunny Greenfelder
Bunny GreenfelderLv2
28 Sep 2019

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Related questions

Strong base is dissolved in 715 mL of 0.400 M weak acid (Ka = 4.38 × 10-5) to make a buffer with a pH of 4.01. Assume that the volume remains constant when the base is added. HA(aq) + OH- (aq) -->H2O(l) + A-(aq)
A.) Strong base is dissolved in 715 mL of 0.400 M weak acid (Ka = 4.38 × 10-5) to make a buffer with a pH of 4.01. Assume that the volume remains constant when the base is added. HA(aq) + OH- (aq) -->H2O(l) + A-(aq)
A. )I found the PKA, which is 4.36 and the mol of HA is .286.
B.)When the reaction is complete, what is the concentration ratio of conjugate base to acid? [A-]/[HA]=?
C.)How many moles of strong base were initially added? moL[OH-]= ?

± pH Changes in Buffers

When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist change in pH following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base (A−) is represented as

HA(aq)⇌H+(aq)+A−(aq)

The buffer will follow Le Châtelier's principle. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and A−. In both instances, [H+] tends to remain constant.

The pH of a buffer is calculated by using the Henderson-Hasselbalch equation:

pH=pKa+log[A−][HA]

Part A

What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.609 mol of NaAin 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.

Express the pH numerically to three decimal places.

Hints

pH =

6.424

SubmitMy AnswersGive Up

Correct

Since both the acid and base exist in the same volume, we can skip the concentration calculations and use the number of moles in the Henderson-Hasselbalch equation to calculate the pH. The answer will be the same.

Part B

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

Express the pH numerically to three decimal places.

Hints

pH =

6.165

All attempts used; correct answer displayed

Even though a significant amount of a strong acid was added, the pH of the buffer changed very little. If this amount of acid was added to 2.00 L of pure water, the pH would change from 7.000 to 1.125.

Part C

What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Express the pH numerically to three decimal places.
pH =

6.408

Incorrect; Try Again; 4 attempts remaining

A strong base is dissolved in 595mL of 0.400M weak acid (Ka =4.16x10^-5) to make a buffer with a pH of 4.02

Assume that the volume remains constant when the base isadded.

HA(aq)+OH-(aq) ---> H2O(l) + A- (aq)


Calculate the pKa value of the acid and determine the number ofmoles of acid initially present.

pKa= ???
??? mol HA


When the reaction is complete, what is the concentration ratio ofconjugate base to acid?
[A- / HA] = ???


How many moles of strong base were initially added?

??? mol OH-

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