Question

Calculate the concentrations of all the ions present in 42.4g K3PO4 in 250.0 mL solution?

Answer 1

`1.93 xx 10^24` ions.
By Type we have: `PO_4 = 4.8 xx 10^23` ions
and `PO_4 = 1.45 xx 10^24` ions

Explanation:

The "ions" would be the individual parts of the compound upon dissociation. Calculate the molar concentration of the original, then use the number of resultant ions to find the "molar concentration" of the ions. Finally, use Avogadro's number to calculate the number of individual ions.

`K_3PO_4` has a molecular weight of `212.3g/"mol"`.

Therefore, `42.4/212.3 = 0.2`moles
`(0.2"mol")/(0.25L) = 0.80M`

This splits into four ions: `K_3PO_4 -> 3K^+ + PO_4`

Therefore, we have a total of `0.80 xx 4 = 3.2` moles of ions.

`3.2 "moles" xx 6.022 xx 10^23 "units"/"mole" = 1.93 xx 10^24` ions.
By Type we have: `PO_4 = 4.8 xx 10^23` ions
and `PO_4 = 1.45 xx 10^24` ions