according to the henderson-hasselbalch equation, if the concentration of a weak acid and its conjugate base are equal, what would be the ph of the solution?
According to the Henderson-Hasselbalch equation, if the concentration of a weak acid and its conjugate base are equal, what would be the pH of the solution?
Answer:
The Henderson-Hasselbalch equation is a useful formula for estimating the pH of a buffer solution. It is expressed as:
\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
where:
- \text{pH} is the hydrogen ion concentration of the solution,
- \text{p}K_a is the negative logarithm of the acid dissociation constant K_a,
- [\text{A}^-] represents the concentration of the conjugate base, and
- [\text{HA}] represents the concentration of the weak acid.
Solution By Steps:
-
Set Concentrations Equal:
If the concentrations of the weak acid [\text{HA}] and its conjugate base [\text{A}^-] are equal, this can be expressed as:[\text{A}^-] = [\text{HA}] -
Substitute into the Henderson-Hasselbalch Equation:
Substitute these equal concentrations into the equation:\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)Given that [\text{A}^-] = [\text{HA}], the ratio \frac{[\text{A}^-]}{[\text{HA}]} becomes:
\frac{[\text{A}^-]}{[\text{HA}]} = 1 -
Calculate the Logarithm of 1:
The logarithm of 1 is 0. Thus, the equation simplifies to:\log(1) = 0 -
Simplify the pH Equation:
Substituting 0 for \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right):\text{pH} = \text{p}K_a + 0Therefore,
\text{pH} = \text{p}K_a
Final Answer:
When the concentration of a weak acid and its conjugate base are equal, according to the Henderson-Hasselbalch equation, the pH of the solution equals the \text{p}K_a of the weak acid. Hence, the pH is \boxed{\text{p}K_a}.