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Lelu [443]
1 year ago
6

Suppose there is 1. 00 l of an aqueous buffer containing 60. 0 mmol of acetic acid (pa=4. 76) and 40. 0 mmol of acetate. Calcula

te the ph of this buffer.
Chemistry
1 answer:
8_murik_8 [283]1 year ago
6 0

For 1. 00 l of an aqueous buffer containing 60. 0 mmol of acetic acid (pa=4. 76) and 40. 0 mmol of acetate, the pH of this buffer is 4.58.

<h3>What is handerson Hasselbalch equation? </h3>

It is expressed as:

pH = pKa + log [A-]/[HA]

where,

[A-] is the molar concentration of conjugate base

[HA] is the molar concentration of weak acid

Given,

pKa = 4.76

<h3>Calculation of concentration:</h3>
  • For acetic acid

C = n/V

= 60/1000

= 0.06M

  • For acetate

C = 40/1000

= 0.04M

Now, substituting values in equation we get,

pH = 4.76 + log(0.04/0.06)

pH = 4.76 + (-0.176)

pH = 4.58

Thus by using Henderson Hasselbalch equation we find the value of pH of the buffer is 4.58.

learn more about pH :

brainly.com/question/9529394

#SPJ4

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pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

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a. If we add 250 mg NaOH (0.250 g)

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b. If we add 350 mg KOH (0.350 g)

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# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

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# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

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