Essay on Acid Base Titration Lab

Submitted By kahoff
Words: 1497
Pages: 6

LAB REPORT

Kim Hoff
Damien Guilband
Noah Jabusch
03/09/15 ­ 3/10/2015
Acid­Base Titration Lab

INTRODUCTION
:
The purpose of this acid­base titration lab was to determine the molar concentration of two acid solutions by performing titrations with a strong base, NaOH, of known concentration. The pH of the solution was monitored and graphed with a pH meter over the course of each titration. The pH graph was used to find the equivalence points of both reactions, allowing the calculation of the concentrations of both acids. The K value for acetic acid was also calculated, using the a
Henderson­Hasselbach equation and the halfway point that was identified on the pH graph.

CHEMICAL EQUATIONS: HCl (aq) + NaOH(aq)

NaCl(aq) + H
O (l)
2
­
­
OH (aq) + HC
H
O (aq) → C
H
O (aq) + H
O (l)
2
3
2
2
3
2
2

MATHEMATICAL FORMULAS:
M = concentration (moles/liter)

M =n÷V

n = amount of solute (moles)
V = volume of solution (liters)

|
Percent error = ||accepted value − experimental value accepted value
| × 100

pH = negative logarithm of hydrogen concentration
­
pH = pK + log([A
]
/
[HA])
a

pK = ­log(K
)
a a K
= equilibrium constant of a weak acid a ­
[HA] and [A
] = concentrations of a weak acid and its conjugate base, respectively.

HAZARDS:
Chemical Name

Formula Properties

Hydrochloric acid

HCl

Sodium Hydroxide NaOH
Acetic acid

Health Hazards

colorless, strong odor, acidic pH Skin/Eye burns or irritation colorless, odorless, basic pH

HC
H
O
2
3
2 colorless with vinegar odor

Skin/Eye burns or irritation
Skin/Eye irritant

PRE­LAB QUESTIONS


1mol KHP × 1mol HP × 1mol NaOH ×
1
1)
0.685g
KHP × 204.22 g KHP


0.181
M NaOH
1mol KHP
.0185L NaOH =
1mol HP −
1mmol CH 3NH 2
1mL CH 3NH 2
2) 15 mL HCl × .2mmol HCl
× .175mmol CH

=
17.1 mL CH
3NH

2
1mL HCl ×
1mmol HCl
3NH 2

QUALITATIVE DATA:
The solutions used in this lab were all originally colorless. As the NaOH solution was added to the acetic and hydrochloric acids, no changes in either solution were observed. RAW QUANTITATIVE DATA:
Hydrochloric Acid titration Volume of
NaOH (mL) 1.95 9.11 10.18 11.18 12.00 12.21 12.34 12.50 12.60 12.67 12.79 12.79 12.82 pH 2.32 2.75 2.91 3.10 3.40 3.51 3.67 3.85 4.06 4.23 4.46 4.80 5.60 dpH/dV 0.06 0.08 0.15 0.26 0.38 0.71 1.15 1.53 2.08 2.51 5.48 28.11 14.94
2
2 d pH/dV
0.005 0.01 0.07 0.15 0.45 1.83 2.81 3.69 7.07 42.47 161.7 230.3 10.45 12.90 12.92 12.96 13.00 13.05 13.29 13.40 13.61 13.80 14.08 14.62 16.22 18.43
6.26 6.83 8.95 9.67 10.08 10.72 10.97 11.20 11.32 11.47 11.67 11.96 12.15
16.34 37.39 32.21 11.90 3.94 2.54 1.56 0.90 0.57 0.38 0.21 0.12 0.08
157.7 170.7 ­240.0 ­240.8 ­44.73 ­10.53 ­4.90 ­2.56 ­1.11 ­0.39 ­0.12 ­0.04 ­0.02 Acetic Acid titration
Volume of
NaOH (mL) 1.10 1.93 3.29 5.12 6.10 7.35 10.38 11.04 11.90 12.10 pH 3.73 4.15 4.56 4.87 5.01 5.21 5.74 5.89
6.18 6.31 dpH/dV 0.47 0.35 0.24 0.17 0.15 0.17 0.19 0.27
0.39 0.59
2
2 d pH/dV ­0.13 ­0.09 ­0.06 ­0.03 ­0.003 0.01 0.03 0.12
0.28 0.51 12.49 12.70 12.90 13.01 13.23 13.40 13.60
6.56 6.71 6.97 7.35 8.22 9.48 10.11
0.70 1.06 2.14 3.84 5.12 4.95 3.80
1.09 3.52 7.99 7.89 2.68 ­2.42 ­5.04

CALCULATED QUANTITATIVE DATA:

With hydrochloric acid

With acetic acid

Moles of NaOH used

0.0011

.0012

Concentration (mol/L)

0.11

0.12

Percent error (concentration)

10.0%

20.0%

K a –

­6
9.8 × 10

Percent error (K
)
a



45.7%

SAMPLE CALCULATIONS:
Moles of NaOH
M = n÷V
0.10 M = n ÷ .01099 L, n = 0.0011 mol Concentration of acid
M = n ÷ V , 1 mol NaOH:1 mol HCl
.0011 mol NaOH: .0011 mol HCl
M = .0011 mol ÷ .010L = .11M

Percent error
|
Percent error = ||accepted value − experimental…