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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?

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

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    $\begingroup$ Because concentration shift to reach equilibrium is negligible. $\endgroup$
    – Poutnik
    Commented Aug 20, 2023 at 5:10
  • $\begingroup$ 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. $\endgroup$
    – Poutnik
    Commented Aug 20, 2023 at 5:11
  • $\begingroup$ 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. $\endgroup$
    – MaxW
    Commented Aug 20, 2023 at 7:15
  • $\begingroup$ Related: chemistry.stackexchange.com/a/172402 $\endgroup$
    – Karsten ♦
    Commented Aug 20, 2023 at 12:16

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