Potentiometric Titration Curves

Data, Results, Calculations and Discussions

Preparation of 250-mL 0.1 M NaOH Solution:

Wt. of NaOH= (vol. of Sol’n) (M of Sol’n) (MW of NaOH)

= (250 mL) (0.1 M NaoH) (40.0g/mol NaOH)

Wt. of NaOH= 1.00 g

• One gram of NaOH pellets was weighed and dissolved in distilled water. The solution was diluted to 250 mL.

Table 1.Weighing of KHP (weighing by difference)

|Replicate |Wt. of container -sample, g |Wt. of KHP |
| |Initial |Final | |
|1 |99.1667 |98.9083 |0.2584 |
|2 |98.9083 |98.6521 |0.2562 |
|3 |98.6521 |98.4016 |0.2505 |

Table 2.Standardization of NaOH

| |Trial 1 |Trial 2 |Trial 3 |
|Initial vol. of NaOH, mL |0.00 |12.80 |25.80 |
|Final vol. of NaOH, mL |12.80 |25.80 |38.90 |
|Volume of NaOH, mL |12.80 |13.00 |13.10 |

Sample calculation for the true concentration of NaOH:

Trial 1

[pic]

=[pic]= 0.09886 M

Table 3.Standardized Concentration of NaOH

|Trial |Concentration of NaOH, M |
|1 |0.09886