Question

1. add 2 to 3 drops of phenolphthalein indicator to the erlenmeyer flask.
2. swirl the flask in order to stir the solution in the erlenmeyer flask for approximately 2 to 5 minutes until all of the acid power has dissolved.
3. use a funnel to fill a clean buret to the 25 ml mark with 0.100 m naoh (sodium hydroxide)
4. place the buret in a buret clamp on the buret stand.
5. record the concentration of the naoh (0.100 m naoh) in the data table.
6. record the initial level of naoh that is in the buret in the data table.
7. place the erlenmeyer flask in the space below the buret so that the naoh can be added.
8. begin the titration by slowly opening the stopcock on the buret so that the naoh is added at a rate of one drop per second. as you titrate, make sure to swirl the flask to stir the contents.
9. continue the titration until the solutions turns a permanent faint pink color and then stop the titration by closing the stopcock on the buret.
10. record the final level of naoh that is in the buret in the data table.
11. if instructed to do a second trial, repeat steps 1 through 13 and record the data in the appropriate places for trial 2 in the data table.

data table

	trial 1	trial 2
concentration of base (naoh)	0.100m
initial level of naoh	25ml
final level of naoh	45.3ml	21.0ml
volume of base used	20.3ml	21.0ml
molar mass of unknown acid	75.0g	71.4

| average molar mass of unknown acid |

data analysis

1. we will first find the number of moles of base (naoh) used to neutralize the unknown acid by using the molarity equation:

molarity = \\(\frac{moles of base}{volume of base used (l)}\\)
na = 23.
o = 16
h = 1

Explanation:

Step1: Convert volume of base to liters for Trial 1

For Trial 1, volume of base used $V_1 = 20.0\ ml=0.0200\ L$, concentration of base $M = 0.100\ M$. Using the formula $n = M\times V$, the moles of base $n_1=0.100\ mol/L\times0.0200\ L = 0.00200\ mol$.

Step2: Calculate molar - mass for Trial 1

Mass of unknown acid $m_1 = 0.15\ g$. Molar - mass $MM_1=\frac{m_1}{n_1}=\frac{0.15\ g}{0.00200\ mol}=75.0\ g/mol$.

Step3: Convert volume of base to liters for Trial 2

For Trial 2, volume of base used $V_2 = 21.0\ ml = 0.0210\ L$, concentration of base $M = 0.100\ M$. Using the formula $n = M\times V$, the moles of base $n_2=0.100\ mol/L\times0.0210\ L=0.00210\ mol$.

Step4: Calculate molar - mass for Trial 2

Mass of unknown acid $m_2 = 0.15\ g$. Molar - mass $MM_2=\frac{m_2}{n_2}=\frac{0.15\ g}{0.00210\ mol}\approx71.4\ g/mol$.

Step5: Calculate average molar - mass

Average molar - mass $MM_{avg}=\frac{75.0\ g/mol + 71.4\ g/mol}{2}=73.2\ g/mol$

Answer:

$73.2\ g/mol$