The aim of this experiment is to standardize 0.1N Sodium Hydroxide (NaOH) which is an unstandard substance, by using standardized Hydrochloric acid (Na2CO3).

1. Soran University
Faculty of Engineering
Department of Petroleum Engineering
Chemistry
Title: Standardization of 0.1N (NaOH) by Standardized (HCl)
Experiment Number: 5
Name: Raboon Redar Mohammed
Group: A
Date: 28/04/2019
Supervisor(s): Dr. Hemn Abdulqadr
Ms. Fenk Abdulrazaq

2. Aim:
The aim of this experiment is to standardize 0.1N Sodium Hydroxide (NaOH) which is
an unstandard substance, by using standardized Hydrochloric acid (Na2CO3).
Introduction:
At the first step, an amount of (NaOH) which also contains Co3 is needed to dissolve
it into 250ml of distilled water to get a "0.1N NaOH solution". Same thing at the
other hand, you need to find N1, and then by the N1V1 = N2V2 theory, you get the
amount of concentrated HCl in (ml) that is needed to obtain "0.1N HCl in 250ml of
distilled water" into a Conical Flask. Dilute 10ml of the 0.1N NaOH into a conical flask
and drop 2 - 3 drops of methyl red using a pipette until its color becomes yellow. Fill
a 250ml burette by the 0.1N HCl and slowly start opening the valve to drop the HCl
into the 10ml 0.1N NaOH conical flask until its color changes to red then close the
valve and stop dropping. Measure the HCl volume remaining in the graduated
cylinder to use it in the N1V1 (HCl) = N2V2 (NaOH) and calculate the N2 of the NaOH.
Materials:
1. Electronic Balance
2. Scoopula
3. Beaker
4. Conical Flask
5. Pipette
6. Methyl Red
7. Burette with Valve
8. Hot Plate
Procedure:
By knowing the Normality (0.1N), Volume (250ml) and equivalent weight (53)...
calculated, 0.95g of NaOH weight is needed to dissolve into a 250ml distilled water to

3. obtain "0.1N NaOH solution". NaOH grains are big, so a hot plain may be needed to
dissolve NaOH in water. By knowing the average, specific gravity, Normality of the
HCl before dilution is calculated which is equal to 12eq/L and then by equaling the
no. of milliequivalence before dilution with the no. of milliequivalence after dilution,
Volume of HCl before dilution is calculated which is equal to 2ml. Now take 2ml of
concentrated HCl with a pipette and dilute it to 250ml with distilled water in a 250ml
Pipette to obtain approximately 0.1N HCl. Drop 10ml of NaOH into a conical flask,
drop 2 - 3 drops of methyl red into conical flask to change its color to yellow. Start
opening the valve to drop the HCl into the 10ml 0.1N NaOH conical flask until its
color changes to red then close the valve and stop dropping. Measure the remaining
HCl volume in the pipette (6ml).
Calculation:
𝑊𝑊𝑊𝑊 =
𝑁𝑁×𝑉𝑉(𝑚𝑚𝑚𝑚)×𝑒𝑒𝑒𝑒.𝑤𝑤𝑤𝑤
1000
=
0.1×250×40
1000
≅ 0.95𝑔𝑔 of Na2CO3 in 250ml of distilled water to
obtain 0.1N Na2CO3.
𝑁𝑁 =
𝑠𝑠𝑠𝑠. 𝑔𝑔𝑔𝑔 × % × 1000
𝑒𝑒𝑒𝑒. 𝑤𝑤𝑤𝑤
=
1.19 ×
37
10
× 1000
36.5
= 12𝑒𝑒𝑒𝑒/𝐿𝐿
𝑁𝑁𝑁𝑁(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. ) = 𝑁𝑁
̀ 𝑉𝑉
̀ (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑. ) → 12 × 𝑉𝑉 = 0.1 × 250 → 𝑉𝑉 =
0.1×250
12
≈ 2𝑚𝑚𝑚𝑚 Take 2ml of
concentrated HCl with a pipette and dilute to 250ml with distilled water in a 250ml
Pipette to obtain approximately 0.1N HCl.
𝑁𝑁𝑁𝑁(𝑁𝑁𝑁𝑁2𝐶𝐶𝐶𝐶3) = 𝑁𝑁
̀ 𝑉𝑉
̀ (𝐻𝐻𝐻𝐻𝐻𝐻) → 0.1 × 6 = 𝑁𝑁
̀ × 10 → 𝑁𝑁
̀ =
0.1 ×6
10
= 0.06𝑁𝑁
Discussion:
One thing to focus on is whenever you see the yellow color changes to red,
immediately close the valve. When it's said yellow to red, you may think that the
color isn't yellow enough and do not close the valve, but actually at the drop the
yellow color changes to red is enough.
Result:
As a result, after the procedure is done on the conical flask, 0.1N standardized NaOH
is obtained.