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Pavel [41]
3 years ago
11

Which of the following is TRUE? A) The equivalence point is where the amount of acid equals the amount of base during any acid-b

ase titration. B) At the equivalence point, the pH is always 7. C) An indicator is not pH sensitive. D) A titration curve is a plot of pH vs. the [base]/[acid] ratio. E) None of the above is true.
Chemistry
2 answers:
liberstina [14]3 years ago
5 0

Answer:

the correct option is B

Explanation:

The correct option is b, since if we reach pH 7, it means that the acid-base reaction is neutralized, therefore the base has been neutralized by an acid or vice versa, without taking into account the proteins or the amounts of both components .

Finger [1]3 years ago
5 0

Answer:

E) None of the above is true

Explanation:

A) is wrong. The unit for amount of substance is the mole. The moles of acid equal the moles of base only when the molar ratio is 1:1. If the molar ratio is different, you will use different moles of acid and base used to reach the equivalence point.

B) is wrong. The pH = 7 only for a strong acid-strong base titration. If you have a weak acid or base, the pH at the equivalence point will not be 7.

C) is wrong. An indicator is pH sensitive. It shows distinct colours in different pH ranges.

D) is wrong. A titration curve is a plot of pH vs. the volume of titrant.

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A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
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Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

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Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

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