Question

If a weak acid is 25% deprotonated at pH 4, what would the pKa be?

Answer 1

Answer:

4.48

pH=pKa+log([A-/HA])

25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.

4 = pKa + log\frac{.25}{.75}

.75

.25

4 - log\frac{.25}{.75}

.75

.25

= pKa

4.48=pKa

Answer 2

Answer:

4.48

pH=pKa+log([A-/HA])

25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.

4 = pKa + log \frac{.25}{.75}

.75

.25

4 - log \frac{.25}{.75}

.75

.25

= pKa

4.48=pKa

Explanation:

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