Question

part a: standardization of a dilute naoh solution.

1. chemical equation for the titration reaction for the standardization of the naoh solution:

khc₈h₄o₄ + naoh = knac₈h₄o₄ + h₂o

1. calculation of the moles of naoh of the closest two trials used.

trial 1
0.000175151
trial 2
0.000 174084

1. calculation of the molarity of naoh for the two trials used.

trial 1
0.000175151
trial 2
0.000174084

1. calculation of the average molarity of the standardized naoh solution for the two trials used.

Explanation:

Step1: Recall molarity formula

The formula for molarity $M=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume in liters. However, since the volume information is not given for each trial to calculate molarity in a proper sense for steps 3, we assume the volume is 1 L for the values given in step 3 (which is incorrect without volume information, but based on the values provided). For step 4, we use the average - formula for two values.

Step2: Calculate average molarity

The average molarity $M_{avg}=\frac{M_1 + M_2}{2}$, where $M_1 = 0.000175151$ and $M_2=0.000174084$.
$M_{avg}=\frac{0.000175151 + 0.000174084}{2}$
$M_{avg}=\frac{0.000349235}{2}= 0.0001746675$

Answer:

$0.0001746675$