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13 Dec 2019

a. What would be the pH of a solution that has equimolar (the same number of moles of) H2PO41- and HPO42- ions? Report your pH to two places past the decimal point.


b. What would the pH be if this buffer solution were made up with 3 times the amount of acid as base?

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Nestor Rutherford
Nestor RutherfordLv2
17 Dec 2019

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Related questions

1.) Calculate the pH of a 0.150 M solution of Na2HPO4. Report just the numerical answer with two digits past the decimal point. Do not include pH = or anything other than the number.

Acid

Ka

H3PO4

7.1 x 10-3

H2PO4-

6.3 x 10-8

HPO42-

4.5 x 10-13

2.) Calculate the pH of a 0.150 M solution of NH4NO3. Report just the numerical answer with two digits past the decimal point. Do not include pH = or anything other than the number. NOTE: this will be the calculated pH on the report form for this week.

Acid

Ka

HNO­3

>>1

NH4+

5.6 x 10-10

3.) Calculate the pH of a 0.150 M solution of KCl. Report just the numerical answer with two digits past the decimal point. Do not include pH = or anything other than the number. NOTE: this will be the calculated pH on the report form for this week.

Acid

Ka

HCl

>>1

A 40.00-mL solution of 0.300 M HOBr (a weak acid with Ka = 2.5x10-9 ) is titrated with a standardized 0.150 M solution of NaOH at 25°C. Enter your numbers to 2 decimal places. What is the pH of the HOBr solution before titrant is added? What is the pH after 40.00 ml total of NaOH has been added? What is the pH after 90.00 ml total of NaOH has been added?

The phosphate buffer system operates in the internal fluid of all cells. H2PO4-(aq) = H+(aq) + HPO42-(aq) Find the pH of this buffer after 0.02 mol of OH- was added to 1.00 l of this buffer containing 0.801 M is H2PO4- and 0.801 HPO42- (Ka1 = 7.2x10-3; Ka2 = 6.1x10-8; Ka3 = 4.2x10-13) assume no volume change due to the addition of 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

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