Question

chem 100l - introduction to chemistry laboratory - experiment #5
data & analysis for titration of vinegar lab - worksheet page 1:
i. standardization
naoh flask code letter = b

	trial 1	trial 2
2 molecular weight of khp (g/mol)	204.23 g/mol	204.23 g/mol
3 moles of khp (divide the value in row one by the value of row 2) show work, with units. sig figs based on row 1.	0.557g / 204.23g/mol =.003mol	.538g / 204.23g/mol =.003mol
4 initial buret volume - final buret volume - ml of naoh used = (all to 2 dec. pl)	0.04 ml 28.34 ml 28.30 ml	.07 ml 27.02 ml 26.95 ml
5 l of naoh used show unit conversion. sig figs based on row 4.	28.30 ml / 1,000 =.028l	26.95 ml / 1,000 =.027l
6 molarity of naoh (divide row 3 by row 5) show work, with units. show to 4 dec. pl.	.003mol /.028l =.1071	.003mol /.027l =.1111

average value of naoh molarity: show working.
7.

Explanation:

Step1: Identify molarity values from previous calculations

Trial 1 molarity of NaOH = 0.1071 M, Trial 2 molarity of NaOH = 0.1111 M

Step2: Calculate the average molarity

The formula for the average of two values \(a\) and \(b\) is \(\text{Average}=\frac{a + b}{2}\). Here \(a = 0.1071\) and \(b=0.1111\). So, \(\text{Average}=\frac{0.1071+ 0.1111}{2}=\frac{0.2182}{2}=0.1091\) M

Answer:

0.1091 M