The Henderson–Hasselbalch equation is an important equation used in chemistry. It helps us to calculate the relationship between the pH of an acid and its dissociation constant (pKa).
A buffer solution is a solution that doesn’t change its pH when an acid or base is added to it. A buffer solution contains acid and its conjugate base, or a base and its conjugate acid—for example, the solution of acetic acid (acid) and sodium acetate (conjugate base).
Using the Henderson–Hasselbalch equation, we can estimate the pH of such a buffer solution.
Henderson–Hasselbalch equation is, therefore, written as follows:
pH = pKa + log [conjugate bases]/[acid]
It is also written as follows:
pH = pKa + log10 ([A–]/[HA])
where,
A- = molar concentration of conjugate base
HA = molar concentration of acid
How is the Henderson–Hasselbalch Equation Derived?
To derive the Henderson–Hasselbalch equation, we need a buffer solution and make various simplifying assumptions.
Assumption 1: We use a monobasic acid that dissociates.
HA -> H+ + A-
Assumption 2: We disregard the self-ionization of water
Assumption 3: We use a salt MA that is completely dissociated in an aqueous solution.
Taking the above assumptions into account, the equation is derived as follows:
HA + H2O ⇋ H+ + A−
The acid dissociation constant (Ka) is added,
Ka = [H+][A−][HA]
Taking, negative log of RHS and LHS:
−log Ka = −log [H+][A−][HA]
⇒−log Ka = −log [H+] – log [A−][HA].
As we know, −log [H+] = pH and −log Ka = pKa,
That is also equal to,
pKa = pH − log [A−][HA]
Rearranging the equation,
⇒pH = pKa + log[A−][HA].
Limitations of this Equation
- This equation does not provide accurate values when strong acids or strong bases are used.
- This equation ignores the self-ionization of water. Hence, the pH of extremely dilute solutions cannot be calculated using this equation.
- We can use this equation to calculate the pH of a polybasic solution. However, the pK value should differ by 3.
Henderson–Hasselbalch Equation in Class 11
This equation is a part of the chapter ‘Acids, Bases, and Salts’ in Class 11. This chapter discusses the different types of acids and bases and the reactions they undergo.
The chapter carries a weightage of 3-4 marks in the exam. Hence, one question based on this equation may be asked.
Illustrated Examples
- Calculate the pH of a buffer that has 0.1M acetic acid and 0.6M acetate. The Ka is 1.8 x 10⁻⁵.
Sol.
We know, pKa = – log 1.8 x 10⁻⁵ = 4.7
Therefore,
pH = pKa + log ([A–]/[HA])
pH = 4.7 + log ( 0.6 / 0.1)
pH = 4.7 + log 6
pH = 4.7 + 0.78
pH = 5.48.
- A buffer solution is made from 0.4M acetic acid and 0.6M acetate. The Ka is 1.8*10⁻⁵, what is the pH of the buffer solution?
Sol:
pH = pKa + log ([A–]/[HA])
pH = 4.7 + log ( 0.6 / 0.4)
pH = 4.7 + log 1.5
pH = 4.7 + 0.17
pH = 4.87.
- A buffer solution is made from 0.2M acetic acid and 0.5M acetate. The Ka is 1.8*10⁻⁵, what is the pH of the buffer solution?
Sol:
pH = pKa + log ([A–]/[HA])
pH = 4.7 + log ( 0.5/ 0.2)
pH = 4.7 + log 2.5
pH = 4.7 + 0.398
pH = 5.098.
FAQs on Henderson–Hasselbalch Equation
Q. Who discovered the Henderson–Hasselbalch equation?
Q. What is pKa?
Q. Does a higher pKa value represent strong acid?
Q. What is pH?
Q. Can we use this equation for a non-buffer solution?
News & Updates
Acids, Bases and Salts Exam
Student Forum
Anything you would want to ask experts?Popular Courses After 12th
Exams: BHU UET | KUK Entrance Exam | JMI Entrance Exam
Bachelor of Design in Animation (BDes)
Exams: UCEED | NIFT Entrance Exam | NID Entrance Exam
BA LLB (Bachelor of Arts + Bachelor of Laws)
Exams: CLAT | AILET | LSAT India
Bachelor of Journalism & Mass Communication (BJMC)
Exams: LUACMAT | SRMHCAT | GD Goenka Test