Question

Calcium phosphate will precipitate out of blood plasma when calcium concentration in blood is 9.2mg/dL, and Ksp for calcium diphosphate is 8.64x10^(-13), what minimum concentration of diphospate results in precipitation?

Answer 1

1.6 ×
`10^(-7)`M

Step-by-step explanation:

`K_(sp)` for calcium diphosphate is 8.64x10^(-13)

concentration of the calcium cations = 2.2955 x 10^−3M

`Ca_(2)P_(2) O_(7)_(s)` ⇄
`2Ca_(2)_(aq)` +
`P_(2)O^(4-) _(7)_(aq)`

`K_(sp)` =
`[Ca^(2+)]^(2)`×
`[P_(2) O^(4-)_7]`

converting the concentration of the calcium cations from 9.2mg/dL to moles per liter , we get 2.2955 x
`10^(-3)`M

`[P_(2) O^(4-)_7]` =
`(K_(sp) )/([Ca^(2+)]^(2) )`

`[P_(2) O^(4-)_7]` =
`(8.64 . 10^(-13) )/([2.2955 . 10^(-3)]^(2) )`

=1.6 ×
`10^(-7)`M