# Why is it that in a buffer solution the equilibrium concentrations may be assumed to be the initial concentrations? [duplicate]

Considering the Henderson–Hasselbalch equation,

$$\text{pH} = \text{p}K_a + \lg \frac{[\ce{AcO⁻}]} {[\ce{AcOH}]}$$

$$\text{p}K_a = \lg \frac{[\ce{AcO⁻}][\ce{H⁺}]} {[\ce{AcOH}]}$$

Why are the values for the concentrations often the initial concentration added?

$$\text{pH} = \text{p}K_a + \lg \frac{[\ce{AcO⁻}]_0} {[\ce{AcOH}]_0}$$

• Because concentration shift to reach equilibrium is negligible. Commented Aug 20, 2023 at 5:10
• Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. Commented Aug 20, 2023 at 5:11
• Remember this isn't pure math where $\pi$ is known to billions of digits. Equilibrium constants are only known to 2 or 3 significant figures generally. Book problems will in almost all cases require nothing more than a quadratic equation to solve by making the right chemical assumptions. Solving 5 equations with 5 unknowns isn't problematic for a computer but it is miserable when doing so by hand.
– MaxW
Commented Aug 20, 2023 at 7:15
• – Karsten
Commented Aug 20, 2023 at 12:16