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Tpy6a [65]
2 years ago
9

Imagine you have a 1.0 mL water sample containing 0.10 M hydrochloric acid, HCl. You want to get it to a safe pH of 6. How much

water do you need to add
Chemistry
1 answer:
AVprozaik [17]2 years ago
7 0

Answer:

See explanation below

Explanation:

In this case, HCl is a strong acid, therefore, it dissociates completely in solution.

To know the quantity of water we need to add, we first need to know the concentration of the acid with pH = 6:

[H⁺] = antlog(-pH)

[H⁺] = antlog(-6) = 1x10⁻⁶ M

This means that the concentration is being diluted.

Now, even if we add great quantities of water, and the concentration and volume change, there is one time that do not change despite the quantity of water added; this is the moles. So, all we have to do, is calculate the moles of the acid in the 1 mL of water, and then, the volume of the acid when it's dilluted:

moles HCl = 0.1 * (1/1000) = 1x10⁻⁴ moles

Now that we have the moles, we can calculate the volume which the acid with the lowest concentration has:

V = mol/M

V = 1x10⁻⁴ / 1x10⁻⁶

V = 100 L

This means that we need to add 99.999 mL of water

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Answer:

Explanation:

The relation between equilibrium constant and Ecell is given below .

E⁰cell = (RT / nF ) lnK  , F is faraday constant T is 273 + 25 = 298 K

E⁰cell  =  1.46 - 1.21 = .25 V

n = 2

Putting the values

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2 )

Change in free energy Δ G

Δ G ⁰ = nE⁰ F

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E⁰ = .4 + .83 = 1.23 V

Δ G ⁰= 4 x 1.23 x 96485

= 474706 J / mol

3 )

E⁰cell = (RT / nF ) lnK

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1.78 = 8.314 x 298  lnK / 2 x 96485

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K = 1.62 x 10⁶⁰

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