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Ahat [919]
3 years ago
5

1

Chemistry
1 answer:
harina [27]3 years ago
5 0

Answer:

Explanation:

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Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid
jasenka [17]

Answer: a) The K_a of acetic acid at 25^0C is 1.82\times 10^{-5}

b) The percent dissociation for the solution is 4.27\times 10^{-3}

Explanation:

CH_3COOH\rightarrow CH_3COO^-H^+

 cM              0             0

c-c\alpha        c\alpha          c\alpha

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.10 M and \alpha = ?

Also pH=-log[H^+]

2.87=-log[H^+]  

[H^+]=1.35\times 10^{-3}M

[CH_3COO^-]=1.35\times 10^{-3}M

[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M

Putting in the values we get:

K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}

K_a=1.82\times 10^{-5}

b)  \alpha=\sqrt\frac{K_a}{c}

\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}

\alpha=4.27\times 10^{-5}

\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}

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4 years ago
PH=3 المحلول حمضي او قاعده ​
Dominik [7]

Answer:

去無奈此刻投哦他家

Explanation:

pH=3 المحلول حمضي او قاعده

5 0
3 years ago
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