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Lina20 [59]
3 years ago
9

A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.

Chemistry
1 answer:
inna [77]3 years ago
7 0
PKa= 4.9 therefore ka= 10^-4.9=  1.259x10^-5

ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}

[CH3CH2COO^-]
= 0.05
[CH3CH2COOH]= 0.10

Therefore 1.259x10^-5 = \frac{[H^+][0.05]}{[0.1]}
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore [H^+] =  \frac{(1.259*10^-5)(0.1)}{0.05}

Therefore [H^+]= 2.513*10^-5

pH= -log [H^+] = -log(2.513*10^-5)= 4.59.
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Answer:

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Explanation:

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<em />

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

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<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>

<em />

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