3 years ago

The Law of Superposition states

Chemistry

1 answer:

lora16 [44]3 years ago

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An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?

juin [17]

Answer:

$pKa=3.58$

Explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

$pH=-log([H^+])$

$[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}$

And the percent ionization is:

$\% \ ionization=\frac{[H^+]}{[HA]}*100\%$

We compute the concentration of the acid, HA:

$[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\$

$[HA]=6.03x10^{-4}$

Thus, the Ka is:

$Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}$

So the pKa is:

$pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58$

Regards.

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Given the following unbalanced chemical reaction: As + NaOH Na3AsO3 + H2 What would be the coefficient of the NaOH molecule in t

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Equal amounts of N2 and O2 are added, under certain conditions, to a closed container. Which changes occur in the reverse reacti

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