The aim of this experiment is to standardize Hydrochloric Acid (HCl) which is an unstandard substance, by using standard Sodium Carbonate (Na2CO3).

1. Soran University
Faculty of Engineering
Department of Petroleum Engineering
Chemistry
Title: Preparation and standardization of (0.1)N HCl by Na2CO3
Experiment Number: 4
Name: Raboon Redar Mohammed
Group: A
Date: 14/04/2019
Supervisor(s): Dr. Hemn Abdulqadr
Ms. Fenk Abdulrazaq

2. Aim:
The aim of this experiment is to standardize Hydrochloric Acid (HCl) which is an
unstandard substance, by using standard Sodium Carbonate (Na2CO3).
Introduction:
At the first step, that amount of Na2CO3 weight is needed to dissolve it into 250ml of
distilled water to get a "0.1N Na2CO3 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 Volumetric Flask. Dilute 10ml of the 0.1N HCl into a conical flask and
drop 2 - 3 drops of methyl orange until its color becomes red. Fill a 250ml graduated
cylinder by the 0.1N Na2CO3 and slowly start opening the valve to drop the Na2CO3
into the 10ml 0.1N HCl conical flask until its colour changes to yellow then close the
valve and stop dropping. Measure the Na2CO3 volume remaining in the graduated
cylinder to use it in the N1V1 (Na2CO3) = N2V2 (HCl) and calculate the N2 of the HCl.
Materials:
1. Electronic Balance
2. Scoopula
3. Beaker
4. Conical Flask
5. Pippete
6. Methyl Orange
7. Graduated Cylinder with Valve
Procedure:
By knowing the Normality (0.1N), Volume (250ml) and equivalent weight (53)...
calculated, 1.325g of Na2CO3 weight is needed to dissolve into a 250ml distilled water
to obtain "0.1N Na2CO3 solution". By knowing the average, specific gravity, Normality
of the HCl before dilution is calculated which is equal to 12eq/L and then by equalling
the no. of milliequivalence before dilution with the no. of milliequivalence after
dilution, Volume of HCl before dilution is calculated which is equal to 2.1ml. Now
take 2.1ml of concentrated HCl with a pipette and dilute it to 250ml with distilled
water in a 250ml Volumetric Flask to obtain approximately 0.1N HCl. Drop 10ml of
HCl into a conical flask, drop 2 - 3 drops of methyl orange into conical flask to change
its color to red. start opening the valve to drop the Na2CO3 into the 10ml 0.1N HCl
conical flask until its colour changes to yellow then close the valve and stop
dropping. Measure the remaining Na2CO3 volume in the cylinder (17.1ml).

3. Calculation:
𝑊𝑊𝑊𝑊 =
𝑁𝑁×𝑉𝑉(𝑚𝑚𝑚𝑚)×𝑒𝑒𝑒𝑒.𝑤𝑤𝑤𝑤
1000
=
0.1×250×53
1000
= 1.325𝑔𝑔 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
= 2.1𝑚𝑚𝑚𝑚 Take
2.1ml of concentrated HCl with a pipette, and dilute to 250ml with distilled water in a
250ml-volumetric flask to obtain approximately 0.1N HCl.
𝑁𝑁𝑁𝑁(𝑁𝑁𝑁𝑁2𝐶𝐶𝐶𝐶3) = 𝑁𝑁
̀ 𝑉𝑉
̀ (𝐻𝐻𝐻𝐻𝐻𝐻) → 0.1 × 17.1 = 𝑁𝑁
̀ × 10 → 𝑁𝑁
̀ =
0.1 ×17.1
10
= 0.171𝑁𝑁
Discussion:
One thing to focus on is whenever you see the red color changes to orange,
immediately close the valve. When it's said red to yellow, you may think that the
color isn't yellow enough and do not close the valve, but actually at the drop the red
color changes to orange is enough.
Result:
At a result, after the procedure is done on the conical flask, (27.1ml) of 0.171N
standard HCl is obtained.