Question

See experiment info below

The NaOH used in this lab should have been 0.1M, since you were to make it as such (see Step 15 in lab document). Compare your standardized average molarity of NaOH in this lab to 0.1M. How do the values compare? How close was your NaOH dilution to actually being 0.1M NaOH? Calculate your percent error against the average standardized value from this lab. (You must show all work to receive credit.)

step 15- Prepare an aliquot (fancy chemical term for a portion from a larger whole) of 15.0 mL of 0.1 M NaOH. Notice that you only have 0.2 M NaOH in your lab kit, so you will need to make a dilution, if you didn’t already do so in a previous lab. To do this, very carefully follow the instructions in the “Before You Start” section on p. 17 in the eScience Student Lab Manual. The procedure for making the dilution is under the heading “Sodium Hydroxide” near the middle of the page. Again, if you already made this dilution, DO NOT dilute your dilution! Just use what you already made (0.1 M NaOH)

Mass of Beaker (Step 1; g)	Mass of Beaker + Citric Acid (Step 3; g)	Mass of Citric Acid (Step 1 - Step 3; g)	Mass of Beaker + Citric Acid After Cooling (Step 7; g)	Molarity of Citric Acid Solution (M)
108.7	110.7	2.00	110.74	0.104M

Syringe reading (mL)	Citric Acid Added (mL)	pH trial 1
30	0.0	11.8
29.5	.5	11.5
29	1	11
28.5	1.5	10.7
28	2	10.5
27.5	2.5	10.3
27	3	10
26.5	3.5	9.6
26	4	9.3
25	5	8.9
24	6	8.7
23	7	8.4
22	8	8.2
21	9	8.1
20	10	7.4
19	11	7.2
18	12	6.9
17	13	6.5
16	14	6.2
15	15	5.5

Syringe reading (mL)	Citric Acid Added (mL)	pH trial 2
30	0.0	_11.6___ initial pH
29.5	.5	11.3
29	1	10.2
28.5	1.5	10
28	2	9.7
27.5	2.5	9.3
27	3	8.7
26	4	7.9
25.5	4.5	7.5
25	5.0	7.2
24.5	5.5	6.9
24	6.0	6.4
23.5	6.5	6.3
23	7	6.1
22.5	7.5	6
22	8.0	5.6
21	9.0	5.3
20	10.0	4.7
19	11	4.2
18	12	4
17	13	3.6
16	14	3.5
15	15	3.2

Answer 1

From the info that you have provided we can calculate the concentration of OH, since NaOH is a strong base and it dissociates completely then we can say that the concentration of OH is the same than the concentration of NaOH so in trial 1 you say you have a PH of 11.8 before the addition of citric acid so i understand that at the very beginning you only have NaOH solution so:

PH + POH = 14

POH = 14 - PH = 14 - 11.8 = 2.2

POH = log [OH]

[OH] = 10^-2.2 = 0.0063

For trial 2

You say you have a ph of 11.6 for pure NaOH solution

PH + POH = 14

POH = 14 - PH = 14 - 11.6 = 2.4

POH = log [OH]

[OH] = 10^-2.4 = 0.00398 M

this is your NaOH concentration

A 0.1 M NaOH solution has a ph of

POH = -log (0.1) = 1

PH = 14 - POH = 14 - 1 = 13