Answer:
1, 2, and 3 are true.
Step-by-step explanation:
The Henderson-Hasselbalch equation is:
pH = pka + log₁₀
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
- If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. TRUE
pH = pka + log₁₀
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
If you know pH and pka:
10^(pH-pka) =
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
The ratio will be: 10^(pH-pka)
- At pH = pKa for an acid, [conjugate base] = [acid] in solution. TRUE
pH = pka + log₁₀
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
0 = log₁₀
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
10^0 =
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
1 =
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
As ratio is 1, [conjugate base] = [acid] in solution.
- At pH >> pKa for an acid, the acid will be mostly ionized. TRUE
pH = pka + log₁₀
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]
- At pH << pKa for an acid, the acid will be mostly ionized. FALSE
pH = pka + log₁₀
![([A^-])/([HA])](https://img.qammunity.org/2020/formulas/chemistry/college/awhrxc2bq91k57agvv2u931h8oo4ur1uwf.png)
If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]
I hope it helps!