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BigorU [14]
2 years ago
7

You must make an aqueous solution that will be a good conductor of electricity. You can pick two substances to add to the water.

Pick the two substances that you could add to water to make it the strongest conductor possible. I) water II) sugar III) hydrochloric acid IV) salt
Chemistry
2 answers:
elixir [45]2 years ago
8 0
IV) Salt and likely III) Hydochloric Acid, because it will dissociate in water similar to the salt.
Arte-miy333 [17]2 years ago
6 0
The acid and the salt are the answers
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Suppose that you have 115 mL of a buffer that is 0.460 M in both benzoic acid ( C 6 H 5 COOH ) and its conjugate base ( C 6 H 5
77julia77 [94]

<u>Answer:</u> The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

<u>Explanation:</u>

We are given:

Concentration of buffer = 0.460 M

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})        .....(1)

We are given:

pK_a = negative logarithm of acid dissociation constant of benzoic acid = 4.2

[C_6H_5COOH]=0.460M

[C_6H_5COO^-]=0.460M

pH = ?

Putting values in equation 1, we get:

pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2

When the pH of the buffer changes by 1 unit, the buffering capacity is said to be lost.

pH change for loosing buffer capacity = [4.2 - 1] = 3.2

Calculating the ratio of conjugate base and its acid by using equation 1:

pK_a = 4.2

pH = 3.2

Putting values in equation 1, we get:

3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}        ......(2)

For benzoic acid and its conjugate base:

Molarity of benzoic acid and its conjugate base = 0.460 M

Volume of solution = 115 mL

Putting values in equation 2, we get:

0.460=\frac{\text{Moles of benzoic acid and its conjugate base}\times 1000}{115mL}\\\\\text{Moles of benzoic acid and its conjugate base}=\frac{0.460\times 115}{1000}=0.053mol

The chemical reaction for aniline and HCl follows the equation:

C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-

Let the moles of acid added to carry out the change is 'x' moles

  • Calculating the moles of acid added:

\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434

Calculating the volume of acid added by using equation 2, we get:

Moles of acid added = 0.0434 moles

Molarity of solution = 0.390 M

Putting values in equation 2, we get:

0.390M=\frac{0.0434\times 1000}{\text{Volume of acid}}\\\\\text{Volume of acid}=\frac{0.0434\times 1000}{0.390}=111.3mL

Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

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Answer:

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