91影视

An element is a pure material made up entirely of atoms with the same number of protons in their nuclei, as defined by chemistry. Chemical elements, unlike chemical compounds, cannot be broken down into smaller substances by chemical reactions.

When phosphate is protonated to generate hydrogen phosphate (${\text{HPO}}_{\text{4}}^{\text{2 -}}$) and dihydrogen phosphate (${\text{HPO}}_{\text{4}}^{\text{2 -}}$), acid rain increases the leaching of phosphates, PO, into groundwater (${\text{PO}}_{\text{4}}^{\text{3 -}}$).

${\text{Ca}}_{\text{3}}{\text{(PO}}_{\text{4}}{\text{)}}_{\text{2}}\text{(aq)}鈫�{\text{3Ca}}^{\text{2 +}}{\text{(aq) + 2PO}}_{\text{4}}^{\text{3 -}}\text{(aq)}$

${\text{Ca}}_{\text{3}}{\text{(PO}}_{\text{4}}{\text{)}}_{\text{2}}$has a solubility product constant of$1.2脳{10}^{-29}$in pure water (pH) according to Appendix C, whereas the solubility product constant may be represented as:

$$\begin{gathered} {\text{K}}_{\text{sp}}\text{=}\frac{{{\text{[Ca}}^{\text{2 +}}\text{]}}^{\text{3}}{{\text{[PO}}_{\text{4}}^{\text{3 -}}\text{]}}^{\text{2}}}{{\text{[Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}} \\ {\text{=1.2脳10}}^{-29} \end{gathered}$$

As calcium phosphate is a solid, its concentration is one, hence it may be left out of the equation:

$$\begin{gathered} {\text{K}}_{\text{sp}}{\text{= [Ca}}^{\text{2 +}}{\text{]}}^{\text{3}}{\text{[PO}}_{\text{4}}^{\text{3 -}}{\text{]}}^{\text{2}} \\ {\text{=1.2脳10}}^{-29} \end{gathered}$$

If we assume that the starting concentration of was one and that the ions concentration was , then after the reaction, some concentration was acquired by ions and some concentration was obtained by ions at equilibrium, as shown in Table below:

As a result, the equilibrium${\text{K}}_{\text{sp}}$may be expressed as

$$\begin{gathered} {\text{K}}_{\text{sp}}{\text{= [3x]}}^{\text{3}}{\text{[2x]}}^{\text{2}} \\ {\text{=1.2脳10}}^{-29} \end{gathered}$$

Then solving the x equation as:

localid="1663309379461" $$\begin{gathered} 27脳x^3脳4脳x^2=1.2脳{10}^{-29} \\ 108脳x^5=1.2脳{10}^{-29} \\ x=\sqrt{\frac{1.2脳{10}^{-29}}{108}} \\ x=6.4439脳{10}^{-7} \end{gathered}$$

Therefore, the calcium phosphate has a solubility of $6.4439脳{10}^{-7}$M at pH .

Each time a${\text{H}}^{\text{+}}$ion is gained from phosphate:

${\text{PO}}_{\text{4}}^{\text{3 -}}鈫�{\text{HPO}}_{\text{4}}^{\text{2 -}}鈫�{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}鈫�{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

In light of the ${\text{K}}_{\text{a}}$${\text{K}}_{\text{a}}$values in Appendix C,