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pH of a mixture strong acid / strong base

Strict method calculation

Calculate for example the $pH$ of $100\;mL$ $HCl\; 0.20\; M$ mixed with $100\;mL$ $NaOH\; 0.10\; M$

Equations

$c_{[HCl]}$ $=$ $\frac{0.10\cdot 0.20 }{0.10+0.10}$ $=$ $0.10\; M$ $c_{[NaOH]}$ $=$ $\frac{0.10\cdot 0.10}{0.10+0.10}$ $=$ $0.050\; M$ (1) Electroneutrality: $[Cl^-]$ $+$ $[OH^-]$ $=$ $[H_3O^+]$ $+$ $[Na^+]$ (2) Conservation of matter: $[Cl^-]=0.10$ (3) Conservation of matter: $[Na^+]=0.050$ (4) Ionic product of water: $[H_3O^+][OH^-]$ $=$ $10^{-14}$

Resolution

Elimination of all variables as a function of $[H_3O^+]$ $[H_3O^+]^2$ $+$ $(0.050$ $-$ $0.10)[H_3O^+]$ $-$ $10^{-14}$ $=$ $0$ Solving this →   2nd degree equation, we find: $[H_3O^+]$ $=$ $5.000\cdot 10^{-2} \frac{mol}{L} $ $pH=1.301$