Respuesta :
Use the h-h equation!
pH = pKa + log[In-/HIn]
2.0 = 9.7 + log[In-/HIn]
-7.7 = log[In-/HIn]
10^-7.7 = [In-/HIn]
[In-/HIn] = 1.995 x 10^-8
Hope this helps!
pH = pKa + log[In-/HIn]
2.0 = 9.7 + log[In-/HIn]
-7.7 = log[In-/HIn]
10^-7.7 = [In-/HIn]
[In-/HIn] = 1.995 x 10^-8
Hope this helps!
Answer:
1.9953 × 10⁻⁸
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pH of the indicator as:
pH = pKa + log[In⁻]/[HIn]
Where Ka is the dissociation constant of the acid.
Given, pKa = 9.7
pH = 2.0
So,
2.0 = 9.7 + log[In⁻]/[HIn]
log[In⁻]/[HIn] = - 7.7
[In⁻]/[HIn] = Antilog (-7.7) = 1.9953 × 10⁻⁸