How to calculate molarity from titration? - (molarity from
titration)
Titration is a volumetric analysis method. It is one of the easiest experimental methods and certainly the most prized possession in a chemist’s toolkit.
We can easily find the molar concentration or molarity of a solution by performing titrimetric analysis.
I know you are excited to find out how we can do so. Hence, without any further delay, let’s start reading!
The amount of a chemical substance dissolved in a specific amount of solution is known as its concentration.
If the amount of solute is measured in moles while the volume of solvent is taken in litres, then it is known as the molar concentration or molarity.
Molarity is calculated in mol/L or M.
For instance, to prepare 1 M (molar) HCl solution, we can use equation (i).
Moles (n) = mass/molar mass …………Equation (i)
The molar mass of HCl is 36.458 g/mol. Putting this value into equation (i).
⇒ Mass = moles x molar mass = 1 x 36.458 = 36.458 grams.
The above calculation shows that 1 M HCl solution is prepared by dissolving 36.458 grams of hydrochloric acid in 1 L of distilled water, which is a very concentrated solution that you need to handle with great care.
If the number of moles of solute in a solution is given, we can find its molarity using equation (ii).
n = C x V …………… Equation (ii)
n= no of moles of solute, C = concentration (molarity), V= volume of the solution in litres.
If the volume of the solution is given in milliliters (mL) which is commonly the case, then the volume given in equation (ii) can be divided by 1000 as;
1 L = 1000 mL or 1 mL = 1/1000 L.
Equation (ii) then transforms into equation (iii).
n = (C × V)/1000……………Equation (iii)
This shows that consistency in units is very important. As the molarity of a solution is always determined in moles/liters so the volume must be in liters (L).
What is titration?
Titration is a wet, volumetric analysis method.
It is one of the easiest ways of finding the unknown concentration or molarity of a chemical solution by titrating it against another solution of known concentration.
In a titration experiment, the solution of known concentration is called the titrant. Contrarily, the solution whose concentration is to be determined is referred to as the titrand.
An acid-base titration is a popular titrimetric analysis method. It is performed to determine the unknown molarity of an acidic or basic solution by reacting it with a corresponding base or acid of known concentration.
An acid-base neutralization reaction takes place in acid-base titrations. It is a chemical reaction of an acid with a base that produces salt and water.
An equivalence point is reached as [H+] = [OH–] in the reaction mixture.
Indicators are organic dyes that mark the endpoint of a titration by giving a quick color change near the equivalence point.
For example, methyl orange is red in color in an acidic solution; it changes to yellow in a basic solution.
Conversely, phenolphthalein changes color from colorless to light pink as the pH of the reaction mixture transforms from acidic (< 7) to basic (> 7) pH.
While performing an acid-base titration experiment, the titrant is usually taken in a burette while the titrand or analyte solution is taken in a titration flask, also known as an Erlenmeyer flask. A few drops of the indicator are added to the analyte solution with the help of a dropper.
For example, we have laid down the following steps to lead you through a general titration experiment.
General steps for titrating an unknown HCl solution against 0.1 M NaOH:
(i). A 0.1 mL aqueous solution of NaOH is prepared in a 100 mL volumetric glass flask.
(ii). A 50 mL burette is filled with this NaOH solution of known concentration. The initial burette reading (x1) is recorded by reading the lower meniscus of the solution, as shown in the figure below.
(iii). A specific volume, such as 10 mL (V2), is transferred to a titration flask with the help of a graduated pipette that allows precise volume measurements.
(iv). A few drops of an indicator, such as phenolphthalein, are added to the titration flask. It is then gently swirled to allow a uniform displacement of the indicator.
(v). The 0.1 M NaOH solution is dropwise added from the burette into the analyte mixture present in the titration flask.
(vi). The titration is stopped immediately as the indicator changes color with one drop of solution from the burette. This marks the endpoint of the titration.
(vii). The final burette reading (x2) is recorded at this point.
(viii). The volume of titrant used till the endpoint is then calculated by subtracting x1 from x2. This is known as titre volume V1 = x2 – x1.
(ix). The above experiment is repeated thrice, and an average of the three titre volumes is noted down for the most accurate calculations. Each reading must not ideally vary more than ±1 units from one another.
How to find molarity from titration?
We can find the molarity of the solution of unknown concentration by applying the titration formula given in equation (iv).
⇒ ……………equation (iv)
Where;
M1 = molarity of the titrant (known concentration)
M2 = molarity of the titrand (unknown concentration)
V1 = titre volume
V2 = volume of titrand or analyte solution taken in the titration flask
n1 and n2 denote the number of moles of the two reacting species as per the balanced chemical equation, respectively.
As per the HCl-NaOH titration example discussed in the previous section, thebalanced chemical equation for the acid-base neutralization reaction is:
1 mole of each of HCl and NaOH reacts completely to form NaCl (salt) and H2O (water); therefore, n1 = n2 = 1.
M1 = 0.1 M NaOH solution was used.
M2 = unknown i.e., to be determined.
V1 = titre volume; let’s suppose 5 mL.
V2 = 10 mL
Then substituting all the above data into equation (iv), we get:
Thus, the molarity of HCl in the above example is 0.05 M or 0.05 mol/L.
In this way, we can find the unknown molarity of a chemical solution by performing an acid-base titration.
Come with us through the following solved examples so that you can have a hands-on practice to apply all the concepts taught in this article.
Solved examples for finding molarity from titration
Example # 1: In a titration of sulfuric acid (H2SO4) against sodium hydroxide (NaOH), 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of H2SO4. Calculate the molarity of sulfuric acid.
In this example, NaOH is the titrant while H2SO4 is the analyte or titrand.
The volume of NaOH used to completely neutralize 26.60 mL of H2SO4 is given in the question statement i.e., titre volume = 32.20 mL.
⇒ V1 = 32.20 mL, V2 = 26.60 mL
The balanced equation for the acid-base neutralization reaction between H2SO4 and NaOH is as follows:
As per the above equation, 1 mole H2SO4 completely reacts with 2 mole NaOH.
⇒ n1 = 2, n2 = 1
The molar concentration of NaOH is known i.e., M1 = 0.250 M. The molarity of H2SO4 is unknown i.e., M2 =?
So we can apply equation (iv) to find M2 as shown below:
Result: The molarity of sulfuric acid is 0.151 mol/L.
Example # 2: In a titration of HCl with NaOH, 15.0 mL of NaOH was required to completely neutralize 10 mL of 0.4 M HCl. What is the molarity of NaOH?
In this particular example, HCl is the titrant while NaOH is the analyte or titrand.
The volume of HCl used to completely neutralize 15.0 mL of NaOH is given in the question statement i.e., titre volume = 10 mL.
⇒ V1 = 10 mL, V2 = 15 mL
The balanced chemical equation for the acid-base neutralization reaction between NaOH and HCl is as follows:
As per the above equation, 1 mole NaOH completely reacts with 1 mole HCl.
⇒ n1 = 1, n2 = 1
The molarity of HCl used is known, i.e., M1 = 0.4 M. The molarity of NaOH is unknown i.e., M2 =?
So we can apply equation (iv) to find M2 as shown below:
Result: The molarity of NaOH is determined to be 0.267 mol/L.
Example # 3: Simi is a chemistry student who collected the following data by titrating 10 mL barium hydroxide Ba(OH)2 against 0.016 M HCl solution in her college laboratory.
Use this titration data to help her find the unknown molarity of Ba(OH)2 used in this experiment.
In this example, HCl is the titrant while Ba(OH)2 is the analyte or titrand.
The initial burette reading is subtracted from the final burette reading to find the volume of titrant used in each experiment.
A mean of the three readings is calculated to find the average titre volume.
Mean titre volume = [(8.8) + (8.7) + (8.9)]/3 = 8.8 mL
The volume of HCl used to completely neutralize 10 mL of Ba(OH)2 is thus 8.8 mL.
⇒ V1 = 8.8 mL, V2 = 10 mL
The balanced chemical equation for the acid-base neutralization reaction between HCl and Ba(OH)2 is written as follows:
As per the above equation, 1 mole Ba(OH)2 reacts completely with 2 moles HCl.
⇒ n1 = 2, n2 = 1
The molarity of HCl used by Simi is given in the question statement i.e., M1 = 0.016 M. She wants to find the molar concentration of Ba(OH)2, thus M2 =?
So we can apply equation (iv) to find M2 as shown below:
Result: The molarity of barium hydroxide used in this acid-base titration is 7.04 x 10-3 mol/L.
Example # 4: Calculate the molarity of an acetic acid solution if 34.57 mL of this solution is needed to neutralize 25.19 mL of 0.1025 M sodium hydroxide.
In this example, NaOH is the titrant while acetic acid (CH3COOH) is the analyte or titrand.
The volume of NaOH used to completely neutralize 34.57 mL CH3COOH is given in the question statement i.e., titre volume = 25.19 mL.
⇒ V1 = 25.19 mL, V2 = 34.57 mL
The balanced chemical equation for the acid-base neutralization reaction between CH3COOH and NaOH is shown as follows:
As per the above equation, 1 mole NaOH reacts completely with 1 mole CH3COOH.
⇒ n1 = n2 = 1
The molarity of NaOH used is given in the question statement i.e., M1 = 0.1025 M. The molar concentration of CH3COOH is unknown i.e., M2 =?
So we can apply equation (iv) to find M2 as shown below:
Result: The molarity of acetic acid used in this example is determined to be 0.0747 mol/L.
Example # 5: 20 mL of a nitric acid (HNO3) solution gets completely neutralized by 0.03 L of 0.20 M NaOH via acid-base titration. Find the molarity of the HNO3 solution.
In this example, NaOH is the titrant while HNO3 is the analyte or titrand.
The volume of NaOH used to completely neutralize 20 mL HNO3 is given in the question statement i.e., titre volume = 0.03 L.
An important point to note here is that the titre volume is given in litres while the volume of titrand is in millilitres. However, as we know already that consistency in units is very important while performing titrimetric calculations.
Thus, we first need to convert 0.03 L into mL as follows:
1 L = 1000 mL
0.03 L = 0.03 x 1000 = 30 mL
⇒ V1 = 30 mL, V2 = 20 mL
The balanced chemical equation for the acid-base neutralization reaction between HNO3 and NaOH is as follows:
As per the above equation, 1 mole NaOH reacts completely with 1 mole HNO3.
⇒ n1 = n2 = 1
The molarity of NaOH used is given in the question statement i.e., M1 = 0.20 M. The molar concentration of HNO3 is unknown i.e., M2 =?
So we can easily apply equation (iv) to find M2 as shown below:
Result: The molarity of nitric acid used in this example is determined to be 0.30 mol/L.
Molarity is one way of measuring the concentration of a chemical solution. It is referred to as the moles of solute dissolved per litre of a solution. It is measured in the units of mol/L or M.
What is titration?
Titration is a wet volumetric analysis method. It is performed to find the concentration of an unknown solution using a solution of known concentration.
What is acid-base titration?
An acid-base titration is performed to find the unknown concentration of an acid or a base against a corresponding base or acid of known concentration respectively.
What is a titrant?
Titrant is a chemical substance whose concentration is known while performing a titration experiment. It is usually taken in a burette.
What is a titrand?
Titrand or analyte is a chemical substance whose concentration is to be determined via a titration experiment. It is usually taken in a titration flask, also known as an Erlenmeyer flask.
What is an indicator in an acid-base titration?
An indicator is a chemical substance that gives a visual sign, such as a color change, to mark the endpoint of a titration experiment.
Usually, conjugated organic dyes such as methyl orange, methyl red, phenolphthalein etc, are used as indicators in acid-base titrations.
These absorb visible radiations of the electromagnetic spectrum and exhibit vibrant colors. The color changes quickly with a small change in the pH of the titration mixture.
What is the equivalence point in an acid-base titration?
The equivalence point is the point at which the concentration of hydrogen ions [H+] becomes exactly equal to the concentration of hydroxide ions [OH–] in the titration flask.
It is where the acid is completely neutralized by the base and vice versa.
What is the difference between an endpoint and an equivalence point?
Equivalence point = pH at which [H+] = [OH–] in an acid-base titration.
Endpoint = pH at which indicator gives a color change with a small drop of an acid or base from the burette.
The best choice of an indicator for a titration experiment is the one whose endpoint coincides with the equivalence point of the acid-base titration.
What is titre volume in an acid-base titration?
Titre volume is the volume of an acid or base required to completely neutralize the corresponding base or acid of unknown concentration.
It is usually calculated by finding the difference between the initial and final burette readings in a titration experiment.
What is a titration curve?
A titration curve is a plot of the pH of the analyte mixture in the titration flask versus the volume of titrant added from the burette.
A titration curve can be used to find the pKa of an acid.
How do you find an unknown molarity from a titration experiment?
An unknown molarity can be determined from the data obtained via a titration experiment by applying the formula given below:
Where;
M1 = molarity of the titrant (known)
M2 = molarity of the titrand (unknown)
V1 = titre volume
V2 = volume of titrand or analyte solution taken in the titration flask
n1 and n2 denote the number of moles of the two reacting species as per the balanced chemical equation, respectively.
While applying the above titration formula, consistency in units is very important; if V1 is in mL, then V2 must be in mL. Similarly, both M1 and M2 are concentrations measured in mol/L or M.
Summary
Molarity is one way of measuring the concentration of a chemical solution.
The molarity or molar concentration of a solution refers to the moles of solute dissolved per liter of a solution. Unit: mol/L or M.
Titration is a wet laboratory method that is used to find the unknown concentration (or molarity) of a solution by using another solution of known concentration (titrant).
The unknown molarity of an acid or base can be determined by titrating it against a corresponding base or acid of known molarity in the presence of an indicator.
Analyte solution is taken in a titration flask while the titrant is added dropwise from the burette.
The volume of titrant used to bring the endpoint of the acid-base neutralization reaction is referred to as the titre volume.
The unknown molarity (M2) can be determined from the known molarity (M1) by applying the titration formula: M1V1/n1 = M2V2/n2 and the balanced chemical equation.
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
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……………equation (iv)
