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Mix equal volumes of $0.1~\mathrm{M}$ $\ce{H3PO4}$ and $0.2~\mathrm{M}$ $\ce{KOH}$. What ion remains in the largest concentration at equilibrium?

My approach: suppose we mix $1~\mathrm{L}$ of each solution. For $\ce{H3PO4}$, $K_{a_1}=7.5\cdot10^{-3}$ and $K_{a_2}=6.2\cdot10^{-8}$, so in terms of the dissociation of $\ce{H3PO4}$, the only ions produced in non-neglible quantities are $\ce{H^+}$ and $\ce{H2PO4-}$. Furthermore the quantities of $\ce{H^+}$ and of $\ce{H2PO4-}$ are both less than one mole, since there is less than 1 mole $\ce{H3PO4}$. We have 2 moles $\ce{OH^-}$ from the $\ce{KOH}$, but since of the $\ce{OH^-}$ neutralizes the $\ce{H^+}$, in particular less than a mole $\ce{OH^-}$ is used in this neutralization reaction. Thus at least one mole $\ce{OH^-}$ remains, at most 1 mole $\ce{H2PO4-}$ remains, and all other ions remain in negligible quantities, so the answer is $\ce{OH^-}$.

Apparently, this is wrong - the answer is $\ce{HPO4^2-}$. I don't see how this is possible because we already established that the amount of $\ce{HPO4^2-}$ formed is negligible from the $K_a$ values, and since we are adding a strong base to a weak acid we know the resulting solution will be basic, so we should expect from the beginning that $\ce{OH^-}$ appears in large concentration.

I don't understand what's wrong with my reasoning?

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Actually, the ion with the largest concentration is $\ce {K+}$.

But that is not what they want. Since potassium hydroxide is a strong base, then the concentration of $\ce {OH-}$ starts with 0.2 M (roughly pH of 13.3). To neutralize this, it would take 2 hydrogens from $\ce {H3PO4}$ at the specified concentration to get both concentrations near $10^{-7}$ (pH ~7). So the major anion is $\ce {HPO4^{2-}}$.

You can find the exact concentrations if you need.

Addendum:
Since $K_w = 10^{-14}$, then $\ce {[H^+][OH^{-}]=10^{-14}}$
Adding 1 L of each solution would give 0.2 moles of $\ce {OH^-}$ to react with any liberated hydrogen from the phosphoric acid. Removing one hydrogen from all the phosphoric acid gives 0.1 mole of $\ce {H+}$, which reacts with the hydroxyl, leaving 0.1 mole of $\ce {OH^-}$ (pH of 13.0). Since this is smaller than ${K_{a2}}$, more hydrogen is released from $\ce {H_2PO_4^{-}}$.

Assuming 0.2 moles of hydrogen is released, then there is an equal amount of $\ce {H+}$ and $\ce {OH^-}$ in the solution. To have $K_w = 10^{-14}$, then each of the ions needs to be near $10^{-7}$M.


Calculations:
Find $\ce {[H_2PO_4^{-}]}$:
$$\ce {H+ + HPO4^{2-} <=> H_2PO_4^{-}}$$
$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$7.5\times 10^{-3}=\frac {[H_2PO_4^{-}]}{10^{-7}\times 0.05}$
$[H_2PO_4^{-}]={10^{-7}\times 0.05}\times {7.5\times 10^{-3}}$
$[H_2PO_4^{-}]=3.75\times 10^{-10}$M


Find $\ce {[PO_4^{3-}]}$:
$$\ce {HPO4^{2-} <=> H^+ + PO_4^{3-}}$$
$K_{a3}=\frac {[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$4.5\times 10^{-13}=\frac {10^{-7}\times [PO_4^{3-}]}{0.05}$
$[PO_4^{3-}]=\frac {0.05\times 4.5\times 10^{-13}}{10^{-7}}$
$[PO_4^{3-}]=2.25\times 10^{-7}$

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  • $\begingroup$ I think the calculations in the second part of the answer,like : $$K_{a2}=\frac {{[H^+]}{[HPO_4^{-2}]}}{[H_2PO_4^{-}]} where \ce {[H+]}=10^{-7}M, and \ce {[HPO_4^{2-}]}=0.05$M.$$ $$6.2\times 10^{-8}=\frac {{10^{-7}\times 0.05}}{[H_2PO_4^{-}]}$$ $$[H_2PO_4^{-}]=\frac{10^{-7}\times 0.05} {6.2\times 10^{-8}}$$ $$[H_2PO_4^{-}]=0.081$$M $\endgroup$ – Adnan AL-Amleh Nov 24 '18 at 0:53
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    $\begingroup$ @AdnanAL-Amleh How can you have a concentration of 0.081 M when the phosphate concentration is 0.05 M ? $\endgroup$ – LDC3 Nov 25 '18 at 1:53
  • $\begingroup$ Sorry, thanks for the interesting and correction; this relation confused me: $$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$$ Because I failed in reconciling it with this relation: $$K_\mathrm{a1} = \frac{[\ce{H3O+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.5 \times 10^{−3}$$ $\endgroup$ – Adnan AL-Amleh Nov 25 '18 at 4:30

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