# Why is Ma*Va=Mb*Vb still valid for weak acids/bases? [duplicate]

I asked my chemistry teacher this today and didn't get an answer. It would seem to me that the equation for determining the morality of the unknown in a titration, MaVa = MbVb, would only work if the acid and the base were both strong.

I always interpreted the equation as a way to say (moles H+)=(moles OH-), but if the acid was weak, the moles of H+ would not simply equal MaVa, because the acid wouldn't fully dissociate, and the same for a base and OH-.

Yet the formula still works for both strong and weak acids and bases. Why?

## marked as duplicate by LDC3, Klaus-Dieter Warzecha, Gimelist, Philipp, M.A.R.Mar 14 '15 at 12:38

We normally titrate a weak acid by a strong base, or weak base by a strong acid, to ensure that the reaction of titration is total (quantitative). We don't titrate weak acid by a weak base (and vice-versa). I will calculate the equilibrium constant of the titration of weak acid $\ce{HA}$ by a strong base $\ce{OH-}$: $$\ce{HA + OH- <=>A- + H2O}$$ The equilibrium constant is $$K=10^{(\mathrm{p}K_w -\mathrm{p}K_a)}$$ For common weak acids (acetic acid $\mathrm{p}K_a=4.7$ ): $$K>> 10^3$$ So, the equation of titration (neutralization) must be written:$$\ce{HA + OH- ->A- + H2O}$$ At the equivalence point: $$n(\ce{HA})=n(\ce{OH-})$$