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Consider only neutral $\mathrm{HA}$ added to pure water, at fixed temperature, with ideal dilute thermodynamic activities, water thermodynamic activity approximated by one, and no added salts or other chemical species. Normalize every concentration by $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$:Conservation of acid units and zero net electric charge, respectively, requireThe second equation balances positive hydronium against negative conjugate base and hydroxide. It would need all additional ions if salts or a chemical base were added. Together with acid dissociation constant $K_a=hx/y$ and the self-ionization of water relation $K_w=ho$, these giveThis is exact within the stated ideal model, including depletion and self-ionization of water. Multiplication by $h(K_a+h)$ yields the cubic equationFor positive $K_a,K_w,c$, the unmultiplied left-hand side increases strictly from negative to positive values as $h$ increases from zero to infinity; its derivative is $1+K_w/h^2+K_{a}c/(K_a+h{)}^2>0$. There is therefore exactly one positive root of a polynomial. A bisection method can find it without choosing an unphysical root of a polynomial.

If $K_w/h^2\ll 1$, neglect hydroxide, so $h\approx x$ and the quadratic equation retaining acid depletion givesThe last form avoids subtracting nearly equal numbers. If also $x/c\ll 1$, neglect depletion to obtainCheck both conditions after computing: $x/c$ measures the fraction lost from the initial undissociated weak acid inventory, while $K_w/h^2$ measures the omitted hydroxide contribution to electric charge balance. A $5\%$ depletion criterion is a chosen tolerance, not a universal boundary. Keeping depletion does not cure the neglect of self-ionization of water. These weak acid mass and charge balances also explain why the square root estimate cannot be extended to arbitrarily dilute chemical solutions.