Calculate the pH of a potassium phosphate buffer if the pKa of KH2PO4 is 7.2 and the acid-base ratio is 2:1
I took Chemistry 7 years ago and am in a review for a Biochem course I am takin. I do not remember the basics so any help with how to get started or what equations to use would be appreciated!
The Henderson-Hasselbalch equation is the one to use.
pH = pKa + log (base)/(acid)
If the aicd/base ratio is 2:1, then the base/acid ratio is 1/2; therefore,
pH = 7.2 + log (1/2)
To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this case, the weak acid is KH2PO4 and its conjugate base is K2HPO4. The acid-base ratio of 2:1 tells us that for every 2 moles of acid (KH2PO4), there is 1 mole of base (K2HPO4).
From the acid-base ratio, we can conclude that [HA] (acid concentration) is twice [A-] (base concentration).
Let's assume the initial concentration of the acid (KH2PO4) is C. Then, the initial concentration of the base (K2HPO4) will be 0.5C.
Now, let's substitute the values in the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = 7.2 + log (0.5C / C)
pH = 7.2 + log (0.5)
To get the pH value, we need to find the logarithm of 0.5. Use a calculator or table to calculate the logarithm of 0.5. The result is approximately -0.301.
pH ≈ 7.2 - 0.301
pH ≈ 6.899
Therefore, the pH of the potassium phosphate buffer is approximately 6.899.
Note: In this calculation, we assumed that the concentration of the acid and base in the buffer solution is equal to their initial concentrations. This assumption holds as long as the buffer capacity is not exceeded.
To calculate the pH of a buffer solution, you can use the Henderson-Hasselbalch equation. In this case, the buffer consists of potassium dihydrogen phosphate (KH2PO4) as the acid and its conjugate base, monopotassium phosphate (K2HPO4), as the base.
The Henderson-Hasselbalch equation is:
pH = pKa + log ([base] / [acid])
Here, pKa is the negative logarithm of the acid dissociation constant (Ka) of the acid, [base] is the concentration of the conjugate base, and [acid] is the concentration of the acid.
In this particular buffer, the acid-base ratio is given as 2:1, which means the concentration of the base is twice that of the acid. Let's assume the initial concentration of the acid is 'C', then the concentration of the base will be '2C'.
Substituting these values into the Henderson-Hasselbalch equation:
pH = 7.2 + log (2C / C)
Simplifying further:
pH = 7.2 + log 2
To calculate the value of log 2, you can use a scientific calculator or refer to logarithm tables. The value of log 2 is approximately 0.301.
Therefore:
pH ≈ 7.2 + 0.301
pH ≈ 7.501
So, the pH of the potassium phosphate buffer would be approximately 7.501.