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max2010maxim [7]
3 years ago
10

Of the following solutions, which has the greatest buffering capacity?

Chemistry
1 answer:
Goshia [24]3 years ago
4 0

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[Base]}{[Acid]} )

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. \frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36

b. \frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417

c. \frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868

d. \frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959

Therefore, the d. solution has the best buffering capacity.

Regards.

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If you burn 50.6 g of hydrogen and produce 452 g of water, how much oxygen reacted?
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3 years ago
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical
Katarina [22]

Answer:

empirical formula: H_2SO_4

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: H_2SO_4.

7 0
3 years ago
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