If 3.60 mL of aluminum hydroxide is used to neutralize 43.0mL of 2.5 M phosphoric acid , what is the molarity if the base? TEN POINTS AWARD?

Al(OH)3 + H3PO4 → AlPO4 + 3 H2O

(43.0mL) x (2.5 M H3PO4) x (1 mol Al(OH)3 / 1 mol H3PO4) / (3.60 mL Al(OH)3) = 29.9 M Al(OH)3

That's an absurd answer. Such a concentrated solution of Al(OH)3 cannot exist at any temperature since Al(OH)3 is nearly insoluble. There's something not right about the given numbers in this question. (The given volumes are especially suspicious.)

Phosphoric acid is H3PO4 and Al(OH)3 is aluminum hydroxide. Luckily, one mole of Al(OH)3 neutralizes one mole of H2PO4 so you just need:

M1V1 = M2V2

V1 = 3.6 mL

V2 = 43.0 mL

and

M2 = 2.5 M

Solve for M1

M1 = M2 (V2/V1)