How to calculate pH from molarity and Ka?, (pH from Ka and molarity)
pH refers to the power of hydrogen. It determines the concentration of hydrogen ions (H+) present in an aqueous solution.
Ka stands for acid dissociation constant. A weak acid only partially dissociates to release a limited number of H+ ions in an aqueous solution. Ka is an equilibrium constant, calculated as a ratio of the product of the concentration of products to the product of reactant concentrations.
Molarity is a way of expressing the concentration of an aqueous solution. It is expressed as moles of solute dissolved per litre of the solution.
In all the above concepts, there is a common term. Yes, you guessed it right; it is concentration.
So what is the link between pH, Ka and molarity? If you are given Ka and the molarity of a weak acid, how can you find the pH of its aqueous solution?
Don’t worry because we will teach you all that and much more through this article. So, continue reading!
An acid dissociates to release hydrogen ions (H+) in an aqueous solution. The amount of H+ ions released in the aqueous solution determines the strength of the acid.
The concentration of H+ ions released can be measured against a parameter called pH (power of hydrogen).
The pH of a solution is related to [H+] by the formula given in equation (i).
pH = -log10 [H+] ….Equation (i)
The hydrogen ion concentration of a solution usually varies from 1 to 10-14 g eq./L. When converted into pH, it is represented in numbers from 0 to 14. The greater the acidity of a solution, the higher [H+], so as per equation (i), so it has a lower pH.
What is Ka?
Ka stands for acid dissociation constant.
A weak acid (HA) dissociates to produce H+ and A– ions in water. H+ ions combine with H2O molecules to form H3O+ ions. A– is known as the conjugate base of the acid. HA and A– together are known as a conjugate acid-base pair.
The ionization equilibrium for the dissociation of HA in an aqueous solution can be represented as follows:
The equilibrium constant (Ka) for the above reaction can be represented as the equation (ii)
Ka = \frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}………. Equation (ii)
Where;
[H3O+] = concentration of hydronium ions formed in the aqueous solution
[A–] = concentration of conjugate base of the acid
[HA] = acid concentration at equilibrium
[H2O] = concentration of water
As water concentration stays constant throughout the reaction, while [H3O+] = [H+], i.e., the concentration of H+ ions released in the aqueous solution. So, equation (ii) can be rearranged as equation (iii).
Ka = \frac{[H^{+}][A^{-}]}{[HA]}……….Equation (iii)
The greater the strength of an acid, it undergoes dissociation to a larger extent in the aqueous solution; thus, it possesses a higher Ka value.
What is molarity?
The concentration of the solution can be expressed using a term called ‘molarity’’.
Molarity is defined as the amount of solute that dissolves in a specific volume of solution. For molarity, the amount is in moles, while volume is taken in litres (L).
So molarity refers to the moles of solute dissolved per litre of the solution.
Units used for molarity: mol/L or M.
The number of moles (n) of solute can be determined using equation (iv):
Moles (n) = mass/molar mass …………Equation (iv)
If the number of moles of solute in a solution is given, we can find its molarity using equation (v).
n = C x V …………… Equation (v)
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 millilitres (mL) which is commonly the case, then the volume given in equation (v) can be divided by 1000 as;
1 L = 1000 mL or 1 mL = 1/1000 L.
Equation (v) then transforms into equation (vi).
………………………..Equation (vi)
This shows that consistency in units is very important. As the molarity of a solution is always determined in moles/litres so the volume must be in litres (L).
What is the relationship between pH, Ka and molarity?
As per equation (iii), if an acid dissociates to a large extent, it possesses a higher Ka value which implies that a large concentration of H+ ions exists at equilibrium.
If the concentration of H+ ions is expressed in mol/L, it is then known as molarity.
So high molarity of H+ implies that the pH value calculated as per equation (i) will be lower.
In this way, pH is inversely related to Ka. In contrast, Ka is directly related to molarity.
If we are provided with Ka and molarities in an aqueous solution, we can easily find [H+] using equation (iii).
Once we have [H+], it is a super easy task to find the pH value of the respective solution using equation (i).
The solved examples given below will help you grasp the diverse ways in which we can apply this concept.
Solved examples for finding pH from Ka and molarity
Example # 1: What is the pH of a 0.45 M solution of acetic acid (CH3COOH)? Ka for acetic acid = 1.58 x 10-5.
CH3COOH is a weak acid that partially dissociates to release H+ and CH3COO– ions in an aqueous solution, as shown below.
Equation (iii) defines Ka.
………. Equation (iii)
This equation can be rewritten w.r.t acetic acid as follows.
The value of Ka for acetic acid is given in the question statement. It is also stated that the initial molarity of acetic acid = 0.45 M.
Let us suppose x moles of H+ ions are released in the aqueous solution at equilibrium. So as per the balanced chemical equation shown above, [H+] equilibrium = [CH3COO–] equilibrium
Hence, [H+] equilibrium = [CH3COO–] equilibrium = x
Note:There are two possible answers for the above expression, but we only choose the positive value because x is a concentration that cannot be negative.
As a final step, we need to substitute the value of x i.e., [H+], into equation (i) to find the required pH.
pH = -log10 [H+] ….Equation (i)
pH = -log10 (2.67 x 10-3) = 2.57
Result: The pH of the acetic acid solution given in this example is 2.57.
Example # 2: What is the pH of a phosphoric acid (H3PO4) solution if 0.35 M acid is left undissociated as the ionization reaction reaches equilibrium?
Ka for H3PO 4 = 7.5 x 10-3.
Phosphoric acid (H3PO4) is a triprotic mineral acid. Its complete ionization in water releases 3 H+ ions per H3PO4 molecule.
Therefore, the complete ionization of H3PO4 occurs under special temperature and pressure conditions only. Otherwise, it dissociates to release a single proton as follows.
So we can use the Ka expression for H3PO4.
W.r.t H3PO4, the Ka expression is:
Let’s suppose [H+] = [H2PO4–] = x at equilibrium. The values of Ka and [H3PO4] equilibrium is given in the question statement.
Now substitute this value of [H+] in equation (i) to find the pH of the given solution.
pH = -log10 [H+] ….Equation (i)
pH = -log10 (0.0512) = 1.29
Result: The pH of the phosphoric acid solution under given conditions is 1.3.
Example # 3: What is the pH of a 0. 1 M aqueous solution of hydrochloric acid (HCl)?
Ka for HCl = 1.3 x 106.
HCl is a very strong acid that completely ionizes to yield H+ and Cl– ions in water. The extremely high Ka value for HCl, as compared to the previous two examples, indicates its high acidic strength.
As HCl completely dissociates into H+ ions, so [H+] = [HCl] as the ionization reaction reaches completion.
[HCl] is given in the question statement i.e., 0.1 M. So we can directly plug it into equation (i) to find the pH of the solution as follows.
pH = -log10 [H+] ….Equation (i)
pH = -log10 (0. 1) = 1.0
Result: The pH of the 0.1 M HCl solution is 1.
Pro-tip:Here, there is no need for Ka expression, so the value of Ka is only given to distract the problem solver. So you need to be very careful and vigilant while solving these kinds of questions.
Example # 4: What is the original molarity of a solution of a weak acid, given Ka and pH?
Ka = 3.5 x 10-5 and pH = 5.25.
Now, this example is the reverse of what we have been doing in the previous three examples.
In this case, the pH value for the weak acidic solution is given so we can directly apply equation (i) to find the H+ concentration.
pH = -log10 [H+] ….Equation (i)
Making [H+] the subject of the formula by taking the antilog of pH.
[H+] = 10-pH = 10-5.25
[H+] = 5.62 x 10-6 mol/L.
A weak acid (HA) dissociates to give an equal amount of H+ and A– ions at equilibrium, as per the balanced chemical equation shown below.
[H+] is already determined, so [A–] = [H+] = 5.62 x 10-6 mol/L.
Let’s suppose the original molarity of HA = x then;
[HA]equilibrium = x – (5.62 x 10-6)
Result: The original molarity of the solution is 6.52 x 10-6 mol/L.
pH stands for the power of hydrogen. It measures the concentration of hydrogen ions (H+) released in an aqueous solution.
Greater [H+] in an aqueous solution lower its pH, which indicates strong acidity.
What is Ka?
Ka stands for acid dissociation constant. It determines the extent of ionization of a weak acid in an aqueous solution. The greater the strength of an acid, the higher its Ka value. However, it has a low pKa value.
What is molarity?
Molarity is one way of expressing the concentration of a solution. A solution consists of a solute and a solvent.
The amount of solute (in moles) dissolved per litre of the solution is known as molarity. It is measured in the units of mol/L, mol/dm3 or simply M.
1 M HCl represents 1 molar solution of hydrochloric acid.
What is the formula that relates pH to molarity?
pH is related to the molarity of hydrogen ions using the formula: pH = – log10 [H+]. Making molarity the subject of the formula, this equation can be transformed as: [H+] = 10-pH.
What is the formula that relates Ka to molarity?
A weak acid (HA) partially dissociates to give H+ and A– ions in an aqueous solution where A– is the conjugate base of the acid.
The Ka for the above reaction at equilibrium can be determined as:
Where
[H+] = molarity of hydrogen ions at equilibrium
[A–] = molarity of conjugate base of the acid
[HA] = molarity of the acid left undissociated at equilibrium.
How can we find the pH of a weakly acidic solution if its original molarity and Ka are given?
The following steps can be followed to find pH from Ka and molarity:
Step I: Suppose the concentration of H+ and A– ions is x.
Step II: Drive an expression for the concentration of HA at equilibrium. [HA]equilibrium = Original molarity – x.
Step III: Apply the Ka expression and find the value of x.
Step IV: Substitute this value of x into [H+] in the formula pH = -log10 [H+], and you will get the required pH value.
Summary
pH refers to the power of hydrogen. It measures the concentration of H+ ions present in an aqueous solution.
The greater the strength of an acid, the lower its pH value.
Ka stands for acid dissociation constant. It determines the extent of ionization of a weak acid in an aqueous solution. Greater the Ka value, the higher the strength of an acid.
Molarity denotes concentration i.e., the number of moles of solute dissolved per litre of solution.
The pH of a solution can be determined by applying the formula pH =- log10 [H+], but for it, the concentration or molarity of H+ ions must be known.
The molarity of hydrogen ions can be determined by applying the expression Ka = [H+][A–]/[HA] if the Ka of the acid is given.
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 (vi)
………. Equation (iii)
Where
