Respuesta :
Answer: The value of [tex]K_a[/tex] for the given acid is [tex]3.58\times 10^{-4}[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Initial mass of weak monoprotic acid = 1.00 g
Molar mass of weak monoprotic acid = 180 g/mol
Volume of solution = 300 mL
Putting values in above equation, we get:
[tex]\text{Molarity of weak monoprotic acid}=\frac{1.00\times 1000}{180\times 300}\\\\\text{Molarity of weak monoprotic acid}=0.0185M[/tex]
To calculate the hydrogen ion concentration for given pH of the solution, we use the equation:
[tex]pH=-\log[H^+][/tex]
We are given:
pH = 2.62
Putting values in above equation, we get:
[tex]2.62=-\log[H^+][/tex]
[tex][H^+]=10^{-2.62}=2.40\times 10^{-3}M=0.0024M[/tex]
The chemical equation for the dissociation of weak monoprotic acid (HA) follows:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Initial: 0.0185
At eqllm: 0.0185-x x x
Evaluating the value of 'x'
[tex]\Rightarrow x=0.0024[/tex]
So, equilibrium concentration of HA = (0.0185 - 0.0024) = 0.0161 M
Equilibrium concentration of [tex]A^-[/tex] = x = 0.0024 M
The expression of [tex]K_a[/tex] for above equation follows:
[tex]K_a=\frac{[H^+][A^-]}{[HA]}[/tex]
Putting values in above equation, we get:
[tex]K_a=\frac{(0.0024)\times (0.0024)}{0.0161}=3.58\times 10^{-4}[/tex]
Hence, the value of [tex]K_a[/tex] for the given acid is [tex]3.58\times 10^{-4}[/tex]
