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kodGreya [7K]
2 years ago
9

What will be the ph of a buffer solution with an acid (pka6. 1) that is exactly half as concentrated as its conjugate base?

Chemistry
1 answer:
anygoal [31]2 years ago
7 0

The pH of a buffer solution with acid that (PKA 6. 1) is exactly half as concentrated as its conjugate base is 6.4.

<h3>What is a buffer solution?</h3>

A buffer solution is a solution that has a maintained pH, not basic or not acidic. Its pH changes when acid or base is added to the solution.

We had to figure out the acid's concentration, which is exactly half that of its potential base.

We know that pH = pH_log

We have less than 6.1 pH so this is a conjugated base.

This will equal to 6.1 + log2 = 6.4

Thus, the pH of a buffer solution with acid is 6.4.

To learn more about buffer solutions, refer to the below link:

brainly.com/question/13169083

#SPJ4

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Answer:

Explanation:

A measurement is the action of measuring something, or some amount of stuff. So it is important to measure certain things right, distance, time, and accuracy are all great things to measure. By measuring these things or in other words, by taking these measurements we can better understand the world around us.

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What would cause the equilibrium to shift left in this reaction?
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The given reaction is:

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In this case, if the rate of the forward reaction is increased then more of the reactants get converted into products i.e. concentration of reactants decreases. In order to undo this change, the equilibrium will shift in a direction to produce more reactants i.e. to the left.

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4 years ago
Enough of a monoprotic acid is dissolved in water to produce a 1.35 M solution. The pH of the resulting solution is 2.93. Calcul
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The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that       [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
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How is it calculated?

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1. Shell number, 1 ≤ n

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3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

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