If the solubility is represented as ' $ \mathrm{S} $ ' $ \mathrm{mol} \mathrm{L} $ and the solubility product as $ \mathrm{K}_{\mathrm{sp}} $, the relation between the two for $ \mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4} $ is

Question

If the solubility is represented as ' $ \mathrm{S} $ ' $ \mathrm{mol} \mathrm{L} $ and the solubility product as $ \mathrm{K}_{\mathrm{sp}} $, the relation between the two for $ \mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4} $ is, $ \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 7} $, $ \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 3} $, $ \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{m}}}{6912}\right)^{1 / 7} $, $ \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5} $

MARKS App · Accepted Answer

Zr 3 PO 4 4 → 3 Zr + 4 + 4 PO 4 - 3 Initial S Final 0 3 S 4 S K sp = 3 S 3 4 S 4 K sp = 27 S 3 × 64 S 4 K sp = 6912 S 7 S = K sp 6912 1 / 7

Search any question & find its solution

Looking for more such questions to practice?