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Lubov Fominskaja [6]
3 years ago
14

Calculate the pH of a 2.3 10-3M[OH'] solution.

Chemistry
1 answer:
disa [49]3 years ago
5 0

Answer:

11.33

Explanation:

-log(2.3x10^-3) = 2.67

14-2.67

- Hope this helped! Let me know if you need a further explanation.

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For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
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This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

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Thus, by recalling the Van't Hoff's equation, we can write:

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Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

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\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

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