pH=6.98
Explanation:
This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.
As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.
2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!
At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have
KW=[H3O+]⋅[OH−]=10−14
Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have
[H3O+]=√10−14=10−7M
The pH of pure water will thus be
pH=−log([H3O+])
pH=−log(10−7)=7
Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10
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The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.
Answer:
0.007 M
Explanation:
pH is defined as the negative logarithm of the concentration of hydrogen ions.
Thus,
pH = - log [H⁺]
The expression of the pH of the calculation of weak acid is:-
Where, C is the concentration = ?
Given, pH = 3.45
So, for
,

C = 0.007 M