Answer to Question #288638 in Molecular Physics | Thermodynamics for mono

Question

El pH de una solución de 0.02 M de un ácido débil desconocido es 3.7. ¿cuál es el pKa de este ácido?

Answer 1

Solution

• [H+] = $10^{-pH}$ so since your acid was measured at 3.7
• you place that into your equation for the pH
• $10^{-3.7}$ placing that into your calculator you will get $19.95\times10^{-5}$ .
• Now $Ka =\frac{ [H+][A-]}{[HA]}$
• So we get as $\frac{(19.95\times10^{-5}M)^2}{0.02M}$ after calculating you will get $9.95*10^{-5}$
• $p_{ka}=-log_{10}(19.95\times10^{-5})\\ = 3.7$
• Suppose,
• $[H+]=10^{-3.7}\\=19.95*10^{-5 }M$
• C*α =[H+]
•  
• $So,α =9.83*10^{-3}$
•  
• $Ka = C*α^2/(1-α) =3.1587*10^{-8 } \\pKa =7.50048$