The most effective buffer solutions have equal concentrations of salt and acid (or base). Find the pH of a buffer solution with equimolar concentrations of:

Question

The most effective buffer solutions have equal concentrations of salt and acid (or base). Find the pH of a buffer solution with equimolar concentrations of:

(a) HCOOH and NaCOOH (Ka = 3.5 x 10-4 for HCOOH)

(b)CH3NH2 and CH3NH3Cl (Kb = 3.7 x 10-4 for CH3NH2)

Thank you!

Answer 1

When equimolar concentrations of base and acid are used, the pH is ALWAYS the pKa. Why?

pH = pKa + log (base)/(acid)
But if base = acid, then base/acid = 1 and log 1 = 0 so pH = pKa.
a is pKa = - log Ka
b is pKb = -log Kb,
then pKa + pKb = 14

Answer 2

To find the pH of a buffer solution, we need to consider the acid-base properties of the components of the buffer and their equilibrium constants (Ka or Kb).

For both parts (a) and (b), we have a weak acid and its conjugate base. In a buffer, the weak acid donates hydrogen ions (H+) when the pH rises, while the weak base accepts hydrogen ions (H+) when the pH decreases, thereby maintaining the pH within a narrow range.

(a) Buffer solution with HCOOH and NaCOOH:

Since HCOOH (formic acid) is a weak acid, it dissociates as follows:
HCOOH ⇌ H+ + COO-

We can write the equilibrium expression for the dissociation of HCOOH as:
Ka = [H+][COO-]/[HCOOH]

The concentration of the weak acid HCOOH and its conjugate base COO- are equal because the buffer is equimolar. Hence,
[HCOOH] = [COO-]

Let's assume the concentration of HCOOH (and NaCOOH) is "C". Therefore,
[HCOOH] = C
[COO-] = C

Substituting these values into the equilibrium expression, we get:
Ka = [H+][C]/C
Ka = [H+]

Since [HCOOH] = [COO-] = C, we assume that the concentration of HCOOH is "C" and that of [H+] is also "C". Hence,
Ka = C

To find pH, we can take the negative logarithm (base 10) of the concentration of [H+]. Therefore,
pH = -log10(C)

(b) Buffer solution with CH3NH2 and CH3NH3Cl:

Since CH3NH2 (methylamine) is a weak base, it accepts hydrogen ions (H+) as follows:
CH3NH2 + H+ ⇌ CH3NH3+

We can write the equilibrium expression for the reaction of CH3NH2 with H+ as:
Kb = [CH3NH3+][H+]/[CH3NH2]

Similar to part (a), let's assume the concentration of CH3NH2 (and CH3NH3Cl) is "C". Therefore,
[CH3NH2] = C
[CH3NH3+] = C

Substituting these values into the equilibrium expression, we get:
Kb = [CH3NH3+][C]/C
Kb = [CH3NH3+]

Since [CH3NH2] = [CH3NH3+] = C, we assume the concentration of CH3NH3+ is "C". Hence,
Kb = C

To find the pOH of the solution, we can take the negative logarithm (base 10) of the concentration of [OH-], which is equal to the concentration of [CH3NH2] and [CH3NH3+]. Therefore,
pOH = -log10(C)

Finally, to find the pH, we can use the relationship between pH and pOH:
pH = 14 - pOH

I hope the explanations above have helped you understand how to find the pH of buffer solutions with equimolar concentrations of weak acid and its conjugate base (or weak base and its conjugate acid).