# pH of weak acids

## Equilibrium calculation

By its dissociation, $HB$ $+$ $H_2O$ $\rightleftarrows$ $H_3O^+$ $+$ $B$ an acid, however small, usually brings a lot more hydrogen ions in water that autoprotolysis of water itself. Let's call ${HB}$ the number of moles of $HB$ introduced into the water before it is dissociated. If $V$ is the final volume, the initial (formal) molarity is written: $c_{HB}$ $=$ $\frac{n_{HB}}{V}$ Let's call $x$ the number of moles of hydronium $H_3O^+$ present in the solution when the dissociation equilibrium is achieved. Then: $H_3O^+$ $=$ $\frac{x}{V}$ and $pH$ $=$ $-log(\frac{x}{V})$ Let's calculate $x$: We find in order: (1) $x$ moles $H_3O^+$ at equilibrium (2) $x$ moles $H_3O^+$ prodiuced (3) $x$ moles $HB$ disappeared (4) $n_{HB}-x$ moles $HB$ at equilibrium (5) $x$ moles $B$ prodiuced (6) $x$ moles $B$ at equilibrium The molarities are at equilibrium: $[H_3O^+]$ $=$ $\frac{x}{V}$ $[HB]$ $=$ $\frac{n_{HB}-x}{V}$ $[B]$ $=$ $\frac{x}{V}$ Introducing in the expression of the acidity constant $K_a$: $K_a$ $=$ $\frac{[H_3O^+][B]}{[HB]}$ = $\frac{\frac{x}{V}\frac{x}{V}}{\frac{n_{HB}-x}{V}}$ = $\frac{\frac{x}{V}\frac{x}{V}}{\frac{n_{HB}}{V}-\frac{x}{V}}$ = $\frac{[H_3O^+]^2}{c_{HB}-[H_3O^+]}$ $[H_3O^+]^2$ $+$ $K_a[H_3O^+]$ $-$ $K_ac_{HB}$ $=$ $0$ So, posing $y=[H_3O^+]$:

To determine the $pH$ of a diluted solution of weak acid, - we calculate a reasonable solution of the equation: $y^2$ $+$ $K_a\cdot y$ $-$ $K_a c_{HB}$ $=$ $0$ - we calculate $pH$ $=$ $-log\;y$

Example: Given a weak acid solution $0.10\frac{mol}{L}$ with $K_a$ $=$ $4.00\cdot 10^{-2}$: The equation $y^2$ $+$ $4.0\cdot10^{-2}\cdot y$ $-$ $4.0\cdot10^{-2}\cdot0.10$ $=$ $0$ provides the positive root: $y$ $=$ $8.3\cdot10^{-3}$ and so: $pH$ $=$ $-log\; y$ $=$ $2.1$