Respuesta :
Answer:
The dissociation constant of this acid is [tex]7.85\times 10^{-5}[/tex].
Explanation:
Moles of acid = [tex]\frac{1.50 g}{176 g/mol}=0.008523 mol[/tex]
Moles of sodium hydroxide:
[tex]Moles=Concentration\times volume (L)[/tex]
Moles of sodium hydroxide:
[tex]=0.300 M\times 0.0125 L=0.00375 mol[/tex]
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]HA+NaOH\rightarrow NaA+H_2O[/tex]
Initial: 0.008523 0.00375 0
Final: (0.008523-0.00375) 0 0.00375
Volume of solution = 50.0 mL + 12.5 mL = 62.5 mL = 0.0625 L (Conversion factor: 1 L = 1000 mL)
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[NaA]}{[HA]})[/tex]
We are given:
[tex]pK_a[/tex] = ?
[tex][HA]=\frac{0.008523 mol-0.00375}{0.0625 L}=\frac{0.004773 mol}{0.0625 L}[/tex]
[tex][NaA]=\frac{0.00375 mol}{0.0625 L}[/tex]
pH = 4.00
Putting values in above equation, we get:
[tex]4.00=pK_a+\log (\frac{\frac{0.00375 mol}{0.0625 L}}{\frac{0.004773 mol}{0.0625 L}})\\\\pK_a=4.105[/tex]
[tex]4.105=-\log[K_a][/tex]
[tex]K_a=7.85\times 10^{-5}[/tex]
The dissociation constant of this acid is [tex]7.85\times 10^{-5}[/tex].
