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hoa [83]
4 years ago
13

Which solution below would have the greatest buffering capacity? The solution was prepared from a weak acid and the salt of its

conjugate base. 0.160 M C6H5OHCOOH and 0.160 M C6H5OHCOONa, Ka = 1.05e-3 0.0892 M H2NCH2COOH and 0.0892 M H2NCH2COOK, Ka = 4.50e-3 0.0725 M H2NCH2COOH and 0.0725 M H2NCH2COONa, Ka = 4.50e-3 0.1360 M C6H5OHCOOH and 0.1360 M C6H5OHCOOK, Ka = 1.05e-3
Chemistry
1 answer:
Tasya [4]4 years ago
5 0

Answer:

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

Explanation:

The first mixture

0.160 M C₆H₅OHCOOH and 0.160 M C₆H₅OHCOONa,

Ka = 1.05x 10⁻³ .

The second mixture

0.0892 M H₂NCH₂COOH and 0.0892 M H₂NCH₂COOK,

Ka = 4.5 x 10⁻³ .

The third mixture

.0725 M H₂NCH₂COOH and 0.0725 M H₂NCH₂COONa,

Ka = 4.5 x 10⁻³

fourth mixture

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

In all the mixtures the ratio of acid and its salt are same and equal to one so this ratio will not determine their relative buffering capacity .

Now we know that weak acid has more buffering capacity so mixture having acid of less Ka will have more buffering capacity .

Ka is less if

Ka = 1.05 x 10⁻³ .

Dilute acids have greater buffering capacity

So ultimate answer is

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

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9 grams of hydrogen gas (H2) will SC  Johnson need to react in order to make 1 bottle of Windex.

Explanation:

Balance equation for the formation of ammonia from H2 gas.

N2 + 3H2 ⇒ 2 NH_{3}

Given

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atomic mass of ammonia 17.01 gram/mole

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number of moles = \frac{51}{17.01}

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in 1 bottle of windex there are 3 moles of ammonia 0r 51 grams of ammonia.

From the equation it can be found that:

3 moles of hydrogen reacted to form 2 moles of ammonia

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