I've read that lead phosphate is very slightly soluble and so any addition of protons shifts the equilibrium back to the solvated ions. What I can't understand is why the lead phosphate is more soluble in hydroxide or basic solutions also.

Source- Vogel's qualitative analysis

• Just because it's more soluble in acid does not mean it's insoluble in base. – Zhe Apr 6 '19 at 20:38
• Is there a reason for it to be soluble in a base? Apparently the base would promote precipitation by removing protons from the solution. – Yash Agarwal Apr 6 '19 at 20:38
• In neutral solution, the number of protons is low enough that if the phosphate/phosphoric acid equilibrium were the driving force for solubility, lead phosphate would be insoluble. You're dealing with two separate facts: (1) lead phosphate is sparingly soluble in water, and (2) lead phosphate is more soluble in acid. – Zhe Apr 6 '19 at 21:15
• That's what I'm saying. Being precipitated in a neutral solution ( which happens) and being reversed in an acidic solution both indicate that the said equilibrium is controlling the solvation- this obviously isn't the case with bases- what happens with bases- why don't they promote precipitation- do they LEAD to some other reactions(obviously do)- what other reactions?? – Yash Agarwal Apr 6 '19 at 21:22
• So, I think I was confused by the statement of your question. "Why lead phosphate dissolved in hydroxide" It dissolves in water, too. What you meant, judging by the accepted answer, is why is it more soluble in basic soluble as well as acidic solution. – Zhe Apr 7 '19 at 1:23

$$\ce{3Pb^{2+} + 2PO4^{3-} <=> Pb3(PO4)2}$$
In acidic solution the phosphate ions would be protonated, thus reducing the concentration of $$\ce{PO4^{3-}}$$ and increasing the solubility of $$\ce{Pb^{2+}}$$.
But the precipitation of the phosphate is not lead's only reaction in alkaline solution. Lead can also form a hydroxide precipitate. In strongly alkaline solution lead can also form various soluble hydroxide complexes which would reduce the concentration of $$\ce{Pb^{2+}}$$ thus increasing the solubility of $$\ce{Pb3(PO4)2}$$.