Answer:
C) [tex]k_a=1.3\times 10^{-5}[/tex]
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:-
[tex]pH=-log(\sqrt{k_a\times C})[/tex]
Where, C is the concentration = 0.5 M
Given, pH = 2.94
Moles = 0.100 moles
Volume = 1.00 L
So, [tex]Molarity=\frac{Moles}{Volume}=\frac{0.100}{1.00}\ M=0.100\ M[/tex]
C = 0.100 M
[tex]2.94=-log(\sqrt{k_a\times 0.100})[/tex]
[tex]\log _{10}\left(\sqrt{k_a0.1}\right)=-2.94[/tex]
[tex]\sqrt{0.1}\sqrt{k_a}=\frac{1}{10^{2.94}}[/tex]
[tex]k_a=1.3\times 10^{-5}[/tex]
The Ka for propionic acid will be "[tex]1.3\times 10^{-5}[/tex]".
Given:
Moles,
Volume in liters,
pH,
We know that,
Concentration of propionic acid will be:
→ [tex]C= \frac{moles}{Volume \ in \ liters}[/tex]
By substituting the values, we get
[tex]= \frac{0.1}{1}[/tex]
[tex]= 0.1 \ M[/tex]
Now,
→ [tex]pH = -log[H+][/tex]
For weak acids,
→ [tex][H+] = (Ka.C)^{\frac{1}{2} }[/tex]
→ [tex]pH = -log(Ka.C)^{\frac{1}{2} }[/tex]
[tex](Ka.C)^{\frac{1}{2} } = 10^{-pH}[/tex]
[tex]Ka = \frac{(10^{-pH})^2}{C}[/tex]
[tex]= \frac{(10^{-2.94})^2}{0.1}[/tex]
[tex]= 1.3\times 10^{-5}[/tex]
Thus the approach above is right.
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