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How to determine the equivalents of OH- in a titration curve
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Nancy Lubin
How to determine the equivalents of OH- in a titration curve
Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base.
$$\ce{H3A + OH- -> H2A- + H2O}$$$$\ce{H2A- + OH- -> HA^{2-} + H2O}$$$$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$
When all$\ce{H3A}$ reacts with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached.
Before the first equivalence point is reached, the $\ce{H3A}$ and $\ce{H2A-}$ buffer exists, whose pH is given by the Henderson-Hasselbalch equation.
$$\mathrm{ pH = pK_{a1} + log\frac{\ce[H_2A^-]}{\ce{[H3A]}}}$$
When $\mathrm{pH = pK_{a1}}$, $\ce{[H2A-]}$=$\ce{[H3A]}$ which means the first half-equivalence point is reached, which requires $\pu{0.5 equivalents}$ of $\ce{OH-}$.
Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base.$$\ce{H3A + OH- -> H2A- + H2O}$$$$\ce{H2A- + OH- -> HA^{2-} + H2O}$$$$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$When all$\ce{H3A}$ reacts with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached.
Before the first equivalence point is reached, the $\ce{H3A}$ and $\ce{H2A-}$ buffer exists, whose pH is given by the Henderson-Hasselbalch equation.$$\mathrm{ pH = pK_{a1} + log\frac{\ce[H_2A^-]}{\ce{[H3A]}}}$$When $\mathrm{pH = pK_{a1}}$, $\ce{[H2A-]}$=$\ce{[H3A]}$ which means the first half-equivalence point is reached, which requires $\pu{0.5 equivalents}$ of $\ce{OH-}$.
Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base. $$\ce{H3A + OH- -> H2A- + H2O}$$ $$\ce{H2A- + OH- -> HA^{2-} + H2O}$$ $$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$ When all $\ce{H3A}$ reacts with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached.
Before the first equivalence point is reached, the $\ce{H3A}$ and $\ce{H2A-}$ buffer exists, whose pH is given by the Henderson-Hasselbalch equation. $$\mathrm{ pH = pK_{a1} + log\frac{\ce[H_2A^-]}{\ce{[H3A]}}}$$ When $\mathrm{pH = pK_{a1}}$, $\ce{[H2A-]}$=$\ce{[H3A]}$ which means the first half-equivalence point is reached, which requires $\pu{0.5 equivalents}$ of $\ce{OH-}$.
Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base.$$\ce{H3A + OH- -> H2A- + H2O}$$$$\ce{H2A- + OH- -> HA^{2-} + H2O}$$$$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$When all $\ce{H3A}$ reacts with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached.
Before the first equivalence point is reached, the $\ce{H3A}$ and $\ce{H2A-}$ buffer exists, whose pH is given by the Henderson-Hasselbalch equation.$$\mathrm{ pH = pK_{a1} + log\frac{\ce[H_2A^-]}{\ce{[H3A]}}}$$When $\mathrm{pH = pK_{a1}}$, $\ce{[H2A-]}$=$\ce{[H3A]}$ which means the first half-equivalence point is reached, which requires $\pu{0.5 equivalents}$ of $\ce{OH-}$.
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