Answer is: <span>reversible reaction.
</span>The concentrations<span> of reactants and products in an equilibrium mixture are determined by </span><span>the </span>equilibrium constant K<span>.
</span>When writing chemical equations for reversible reactions, two arrows are used, for example: NH₄Cl(s) ⇆ NH₃(g) + HCl(g) or
CuSO₄(s) + 5H₂O(l) ⇆ CuSO₄·5H₂<span>O(s).</span>
Answer:
[C₆H₅NH₃⁺] = 0.0399 M
Explanation:
This excersise can be easily solved by the Henderson Hasselbach equation
C₆H₅NH₃Cl → C₆H₅NH₃⁺ + Cl⁻
pOH = pKb + log (salt/base)
As we have value of pH, we need to determine the pOH
14 - pH = pOH
pOH = 8.43 (14 - 5.57)
Now we replace data:
pOH = pKb + log ( C₆H₅NH₃⁺/ C₆H₅NH₂ )
8.43 = 9.13 + log ( C₆H₅NH₃⁺ / 0.2 )
-0.7 = log ( C₆H₅NH₃⁺ / 0.2 )
10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2
0.19952 = C₆H₅NH₃⁺ / 0.2
C₆H₅NH₃⁺ = 0.19952 . 0.2 = 0.0399 M
Deleting it and downloading it back again
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Answer:
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