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Nadusha1986 [10]
3 years ago
15

If the Ka of a monoprotic weak acid is 5.4 × 10-6, what is the pH of a 0.25 M solution of this acid?

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0
The first step is to write the balanced equation of the dissociation of the monoprotic weak acid. This can be assumed to have 1:1 stoichiometric relationships, so the equation and ICE chart are given by:

                          HA ---> H^+ + A^-
initial                 0.25       0          0
change               -x         x           x
equilibrium    0.25-x       x           x

This gives an expression for Ka given by:

Ka = (x)^2/(0.25-x) = 5.4x10^-6
x = 1.159x10^-3 = [H+] = [A-]

pH = -log[H+] = -log[1.159x10^-3] = 2.936
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100 g

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From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

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Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

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Thus, 2 half-lives has elapsed.

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N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

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