Question

experiment 3: titration, standardization, and molar mass determination
name: brooklynn fere
date: 9 - 23 - 25 section: 6
directions: complete the following worksheet using your own data collected during lab. all values must include proper units and the correct number of significant figures.
part a: standardization of a dilute naoh solution.

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

kh{c}_{8}{h}_{4}{o}_{4}+naoh = nak{c}_{8}{h}_{4}{o}_{4}+{h}_{2}o

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

trial 1
0.000175151
trial 2
0.000174084

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

trial 1
0.000175151m
trial 2
0.000174084m

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

= 0.0001746675mol

1. calculation of the % difference of the two trials used (to 2 significant digits).

Explanation:

Step1: Recall percentage - difference formula

The formula for percentage difference is $\text{\% Difference}=\frac{\vert\text{Value}_1 - \text{Value}_2\vert}{\frac{\text{Value}_1+\text{Value}_2}{2}}\times100\%$. Let $\text{Value}_1 = 0.000175151$ and $\text{Value}_2=0.000174084$.

Step2: Calculate the numerator

$\vert0.000175151 - 0.000174084\vert=\vert0.000001067\vert = 0.000001067$.

Step3: Calculate the denominator

$\frac{0.000175151 + 0.000174084}{2}=\frac{0.000349235}{2}=0.0001746175$.

Step4: Calculate the percentage - difference

$\text{\% Difference}=\frac{0.000001067}{0.0001746175}\times100\% \approx 0.61\%$.

Answer:

$0.61\%$