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Andrews [41]
3 years ago
8

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

ate the Ka for the acid.
Chemistry
1 answer:
Ede4ka [16]3 years ago
6 0
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:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M
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siniylev [52]

Answer:

10^{-3.4

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Let us first take a look at the image below;

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∴ The equilibrium constant for the acid-base reaction is expressed as:

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6 0
3 years ago
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NNADVOKAT [17]

Answer:

1

Explanation:

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Answer:

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Option b. is the correct one.

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