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kolbaska11 [484]
3 years ago
9

An acid with a p K a of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated f

orm of the acid?
Chemistry
1 answer:
olga2289 [7]3 years ago
7 0

Explanation:

According to the Henderson-Hasselbalch equation,

                  pH = pK_{a} + \frac{log[A^{-}]}{[HA]}

Given values are pH = 6, pK_{a} = 8

Putting given values into the above equation as follows.

                   6 = 8 + \frac{log [A^{-}]}{[HA]}

                   \frac{log[A^{-}]}{[HA]} = -2

                   \frac{[A^{-}]}{[HA]} = antilog -2

                                         = 0.01

But according to the question, we need protonated to deprotonated ratio of \frac{[HA]}{[A^{-}]}

            \frac{[HA]}{[A^{-}]} = \frac{1}{0.01}

                     \frac{[HA]}{[A^{-}]} = 100

Thus, we can conclude that ratio of the protonated to the deprotonated form of the acid is \frac{100}{1}.

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Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
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Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

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Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

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T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

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