Respuesta :
The answer is the phosphate concentration of [tex]8.40\times10^{-6}M[/tex]
Calcium phosphate [tex]$\left(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right)$[/tex] is dissociate to form
calciumion [tex]$\left(\mathrm{Ca}^{2+}\right)$[/tex] and phosphate ion [tex]$\left(\mathrm{PO}_{4}^{3-}\right)$[/tex].
[tex]$\mathrm{Ca}_{2}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s}) \longrightarrow 3 \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})$[/tex]
Soluability product expression of [tex]$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$[/tex] is,
[tex]$K_{s p}=\left[\mathrm{Ca}^{2+}\right]^{3} \times\left[\mathrm{PO}_{4}^{3-}\right]^{2} \longrightarrow (1)[/tex]
given that, somability product [tex]$\left(k_{s p}\right)=8.6 \times 10^{-19}$[/tex]
concentration of [tex]\mathrm{ca}^{2+}$ ion $=0.0023 \mathrm{M}$[/tex]
putting the value in equation (1) we get,
[tex]$8.6 \times 10^{-19}=(0.0023)^{3} \times\left[\mathrm{PO}_{4}^{3-}\right]^{2}$[/tex]
or, [tex]$\left[\mathrm{PO}_{4}^{3-}\right]^{2}=\frac{8.6 \times 10^{-19}}{(0.0023)^{3}}=7.06 \times 10^{-11}$[/tex]
or, [tex]$\left[\mathrm{PO}_{4}^{3-}\right]=8.40 \times 10^{-6} \mathrm{M}$[/tex]
So the phosphate concentration [tex]$=8.40 \times 10^{-6} \mathrm{M}$[/tex]
What is phosphate concentration ?
- Phosphorus is a component of adenosine diphosphate (ADP) and adenosine triphosphate (ATP), both of which utilise the bonds formed between phosphate groups to store energy.
- Although it is also found in plasma, phosphate is a significant intracellular anion.
- Adults' typical serum phosphate concentrations fall between[tex]2.5$ to $4.5 \mathrm{mg} / \mathrm{dL}(0.81$ to $1.45 \mathrm{mmol} / \mathrm{L})$[/tex]
So the more about phosphate concentration visit.
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