Question

Write charge and mass balances for aqueous ${\mathbf{Ca}}_{\mathbf{3}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{2}}$ if the species are ${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}\mathbf{,}{\mathbf{CaOH}}^{\mathbf{+}}\mathbf{,}{\mathbf{CaPO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{,}{\mathbf{PO}}^{\mathbf{3}\mathbf{-}}\mathbf{,}{\mathbf{HPO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{,}{\mathbf{H}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{,}{\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}$.

Short answer

The charge and mass balances for aqueous${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

charge balance.

role="math" localid="1654834418096" $2\left[{\mathrm{Ca}}^{2+}\right]+\left[{\mathrm{CaOH}}^{+}\right]+\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{CaPO}}_4^{-}\right]+3\left[{\mathrm{PO}}_4^{3-}\right]+2\left[{\mathrm{HPO}}_4^{2-}\right]+\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]\left[{\mathrm{OH}}^{-}\right]$

Mass balance

role="math" localid="1654834668451" $$\begin{gathered} \underset{}{\underset{\mathrm{species}\mathrm{containing}\mathrm{calcium}}{2\left\{\underbrace{\left[{\mathrm{Ca}}^{2+}\right]+\left[{\mathrm{CaOH}}^{+}\right]+\left[{\mathrm{CaPO}}_4^{-}\right]}\right\}}} \\ =\underset{\mathrm{species}\mathrm{containing}\mathrm{phosphate}}{\underbrace{3\left\{\left[{\mathrm{CaPO}}_4^{-}\right]+\left[{\mathrm{PO}}_4^{3-}\right]+2\left[{\mathrm{HPO}}_4^{2-}\right]+\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]\right\}}} \end{gathered}$$

Step-by-step solution

Define The charge and mass balance

Chemical equations for mass and charge are frequently balanced in aqueous solutions. On both sides of the equation, balancing for mass yields the same number and types of atoms. When the net charge on both sides of the equation is zero, the equation is said to be balanced for charge.

Mass balance: All species in a solution containing a certain atom (or group of atoms) must be equal to the amount of that atom (or group) given to the solution.

Step 2: Calculate the charge and mass balance.

Given

Aqueous${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

Species:${\mathrm{Ca}}^{2+},{\mathrm{CaOH}}^{+},{\mathrm{CaPO}}_4^{-},{\mathrm{PO}}^{3-},{\mathrm{HPO}}_4^{2-},{\mathrm{H}}_2{\mathrm{PO}}_4^{-},{\mathrm{H}}_3{\mathrm{PO}}_4$

Now charge balance.

$2\left[{\mathrm{Ca}}^{2+}\right]+\left[{\mathrm{CaOH}}^{+}\right]+\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{CaPO}}_4^{-}\right]+3\left[{\mathrm{PO}}_4^{3-}\right]+2\left[{\mathrm{HPO}}_4^{2-}\right]+\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]\left[{\mathrm{OH}}^{-}\right]$

By the definition of charge balancerole="math" localid="1654834648341" $\underset{}{\underset{\mathrm{species}\mathrm{containing}\mathrm{calcium}}{2\left\{\underbrace{\left[{\mathrm{Ca}}^{2+}\right]+\left[{\mathrm{CaOH}}^{+}\right]+\left[{\mathrm{CaPO}}_4^{-}\right]}\right\}}}$

$=\underset{\mathrm{species}\mathrm{containing}\mathrm{phosphate}}{\underbrace{3\left\{\left[{\mathrm{CaPO}}_4^{-}\right]+\left[{\mathrm{PO}}_4^{3-}\right]+2\left[{\mathrm{HPO}}_4^{2-}\right]+\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]\right\}}}$