Question

1) What is the difference between endpoint and equivalence point in a titration? Why is it important to know the difference when analyzing your results? Unknown HCl 1: Trial # 1 Vol HCI (ml) Initial % Error (NaOHT (mol/L) Volfinal- Volinitial = (mL) Total Volume NaOH (ml) 10.00 10.00 10.00 10.00 0.098 0.098 0.098 0.098 Volume NaOH (mL) 5.52 10.29 15.21 22.20 Volume NaOH (mL) 0.00 5.52 10.29 15.21 3 L Average: Unknown HCl 2: Trial # 1 Vol HCI (mL) %E (NaOHT (mol/L) Volfinal - Volinitial = (ml) Total Volume NaOH (mL) Final Volume NaOH (mL) 8.62 16.45 24.09 32.15 Initial Volume NaOH (mL) 0.80 8.62 16.45 24.09 Rapid 10.00 10.098 10.00 0.098 10.00 0.098 10.00 0.098 2 3 L Average: True [HCI] Solutions: Unknown 1: 0.0521M Unknown 2: 0.0811M Lastly, please answer the following questions and show any necessary work in your lab notebook. Once you have completed this worksheet, please send photos or scans of your work to your lab instructor via email. 1. An acid dissociates to form what type of ion? a. OH+ b. OH c. H d. H+ 2. What type of reaction occurs when an acid and base reacts? a. oxidation b. neutralization c. physical d. emotional 3. Dissolving 0.729 mol NaCl in 2530 mL of water yields a solution with what concentration? a. 0.000288 M b. 0.288 M c. 1.84 M d. 3.47 M 4. (TRUE / FALSE ) Liquid water is a product of an acid and base reaction.

Answer 1

1)

Equivalence point gives the exactly how much volume of acid or base required to completely neutralise the given volume of base or acid solution.

End point gives the indication that the neutralisation reaction is completed.

For example when we are titrating HCl Vs NaOH, methyl orange or phenolphthalein indicator is used.

Phenolphthalein indicator shows no colour (colourless) in acidic medium and pink colour in basic medium.

Let us assume 10 mL of 0.1 M HCl is taken in a conical flask and add 2-3 drops of phenolphthalein indicator. After addition of indicator, the solution is colourless. When NaOH is added to HCl, HCl is neutralised. When 9.9 mL of NaoH is added, still the solution is acidic and is colourless only.

When 10 mL of NaOH is added, the solution is neutral and the solution is colourless. It is the equivalence point.

When 10.1 mL is added, the solution becomes basic. Hence colour changes from colourless to pink. It is the end point.

In titrametry of an acid and base, we will get only end point but not equivalence point. Hence the result may have little bit error.

Ttial	Volume of HCl/mL	Molarity of NaOH/M	Final Volume of NaOH /mL	Initial volume of NaOH	Volume final - Volume initial	Total volume of NaOH
1	10.00	0.098	5.52	0	5.52	5.52
2	10.00	0.098	10.29	5.52	4.77	4.77
3	10.00	0.098	15.21	10.29	4.92	4.92
4	10.00	0.098	22.20	15.21	6.99	6.99
Average volume of NaOH	5.55

HCl + NaOH ------> NaCl + H2O

$\frac{M_{1}V_{1}}{n_{1}} = \frac{M_{2}V_{2}}{n_{2}}$

M1 : Molarity of HCl = ?

V1 : Volume of HCl = 10.00 mL

n1 = number of moles of HCl = 1

M2 = Molarity of NaOH = 0.098 M

V2 : Average volume of NaOH = 5.55 mL

n2 : Number of moles of NaOH = 1

M​​​​​​1 = 0.098 M * 5.55 mL *1 mole/(1 mole * 10.00 mL)

M​​​​​​1 = 0.05439 M

Trial 2:

Ttial	Volume of HCl/mL	Molarity of NaOH/M	Final Volume of NaOH /mL	Initial volume of NaOH	Volume final - Volume initial	Total volume of NaOH
1	10.00	0.098	8.62	0.80	7.82	7.82
2	10.00	0.098	16.45	8.62	7.83	7.83
3	10.00	0.098	24.09	16.45	7.64	7.64
4	10.00	0.098	32.15	24.09	8.06	8.06
Average volume of NaOH	7.84

HCl + NaOH ------> NaCl + H2O

$\frac{M_{1}V_{1}}{n_{1}} = \frac{M_{2}V_{2}}{n_{2}}$

M1 : Molarity of HCl = ?

V1 : Volume of HCl = 10.00 mL

n1 = number of moles of HCl = 1

M2 = Molarity of NaOH = 0.098 M

V2 : Average volume of NaOH = 7.84 mL

n2 : Number of moles of NaOH = 1

M​​​​​​1 = 0.098 M * 7.84 mL *1 mole/(1 mole * 10.00 mL)

M​​​​​​1 = 0.0768 M

1) The acid is a substance which gives Hydronium ions (H+) is called an acid

Option D is correct answer.

2) Acid-base reaction is called as a neutralisation reaction.

3)

Molarity is the number of moles of solute present in one litre of solution and is denoted by M

$M = \frac{number of moles of solute *1000}{Volume of solution in mL}$M = 0.729 *1000/2530 mL

M = 0.288 M

Option B is correct answer.

4)

Liquid water is formed when an acid is reacted with a base. Along with liquid water, salt is also formed.

The given statement is True

HCl (aq)+ NaOH (aq) ------> NaCl (aq)+ H2O (l)