Henderson's Hasselbalch equation for the propanoic acid
[tex]solution \: is \: pH=4.874 + log \frac{[A ^{ - } ] }{HA}[/tex]
This is called known as the Henderson- Hasselbalch equation which relates the pH of an acid aqueous solution with the acid dissociation constant.
[tex]pH=pK _{a } + log \frac{Conjugate \: base}{acid}[/tex]
The Henderson-Hasselbalch equation is derived from the definition of the acid dissociation constant as follows.
Consider the hypothetical compound HA in water.
[tex]HA +H _{2}O→ H _{3} O ^{ + } + A^{ - } [/tex]
The acid dissociation constant of HA,
[tex] [H _3O ^{ + }] = K _a \times \frac{HA}{[A ^{ - } ] }[/tex]
[tex]log [H _3O ^{ + }] = logK_a+log\frac{HA}{[A ^{ - } ] }[/tex]
Multiply both sides of the equation by -1.
[tex]-log[H_3O ^{ + }]=-logK_a-log\frac{HA}{[A ^{-} ] }[/tex]
[tex]pH=pK _{a }+log \frac{[A ^{ - } ] }{HA}[/tex]
[tex][HA] = [A-][/tex]
[tex]pH =pKa +log1[/tex]
[tex]pH = pKa + 0[/tex]
pH=pKa
pKa value of compounds can be determined experimentally using its relation with pH.
According to the Henderson-Hasselbalch equation, when the concentrations of the acid and the conjugate base are the same, i.e when the acid is 50% dissociated, the pH of the solution is equal to the pKa of the acid.
[tex]pK _{a } \: value \: for \: propanoic \:acid \: = 4.874[/tex]
[tex]Let \: propanoic \: acid \: be \: HA.[/tex]
[tex]Let \: propionate \: be \:A ^{ - } .[/tex]
So Henderson-hasselbalch equation for the propanoic acid solution is,
[tex]pH=pK _{a } + log \frac{[A ^{ - } ] }{HA}[/tex]
[tex]pH=4.874 + log \frac{[A ^{ - } ] }{HA}[/tex]
Therefore, Henderson's Hasselbalch equation for the propanoic acid solution is
[tex]pH=4.874 + log \frac{[A ^{ - } ] }{HA}.[/tex]
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