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sweet-ann [11.9K]
3 years ago
10

A student had 1.00 L of a 1.00 M acid solution. Much to the surprise of the student, it took 2.00 L of 1.00 M NaOH solution to r

eact completely with the acid. Explain why it took twice as much NaOH to react with all of the acid. In a different experiment, a student had 10.0 mL of 0.020 M HCl. Again, much to the surprise of the student, it took only 5.00 mL of 0.020 M strong base to react completely with the HCl. Explain why it took only half as much strong base to react with all of the HCl.
Chemistry
1 answer:
miss Akunina [59]3 years ago
7 0

Answer:

1. A diprotic acid was neutralized.

2. HCl has the ability to accept two protons.

Explanation:

Hello,

Neutralization chemical reactions are better quantified by using equivalent grams instead of moles since they define the mass of a given substance which will neutralize the other substance, thus, in the first, experiment, 1 mole of NaOH has 1 equivalent grams since there's just one hydroxile in its structure, on the other hand, as 2.00 L of a 1.00 M solution of NaOH were needed to neutralize 1.00 L of a 1.00 M solution of an acid, one concludes that the acid was diprotic, it means that it has two hydrogens in its structe.

In such a way, in the second experiment, it was the other way around but by using the hydrochloric acid as it has the ability to accept two protons, it means that the base had two hydroxile ions in its structure.

Best regards.

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C. The points must be placed one at a time on each side of the chemical symbol: it is correct, because that is the way to make the point diagram.

D. An atom is chemically stable if all the points are paired: this is correct since this verifies that the point diagram has been done well.

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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
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Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

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                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

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Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

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