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3 years ago

If a weak acid is 25% deprotonated at ph 4, what is the pka?if a weak acid is 25% deprotonated at ph 4, what is the pka?

2 answers:

7 0

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 $\frac{.25}{.75}$
4 - log $\frac{.25}{.75}$ = pKa
4.48=pKa

nikitadnepr [17]3 years ago

4 0

Answer:

pKa = 4.5

Explanation:

The weak acid can be represented by the general formula, HA and the dissociation equilibrium given as:

$HA \rightleftharpoons H^{+}+A^{-}$

where HA = protonated form

A- = deprotonated form

The Henderson-Hasselbalch equation relates the pH of a solution to the ratio of the concentrations of HA and A- as;

$pH = pKa + log\frac{[A-]]}{[HA]]}-----(1)$

It is given that:

% deprotonated i.e. A- = 25%

Therefore, %protonated i.e. HA = 100 -25 = 75%

pH = 4

Based on equation (1)

$4 = pKa + log\frac{[25]]}{[75]]} = pKa-0.477$

pKa = 4.477 i.e. around 4.5

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A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

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$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

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Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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