Ascientist completes a titration by adding 36.0 ml of ca(oh)2 (aq) of 1.0 m concentration to 11.0 ml of an unknown h3po4 (aq) solution. what is the molar concentration of the h3po4 (aq)?

We calculate for the number of moles of calcium hydroxide Ca(OH)2 which is equal to number of moles of phosphoric acid H3PO4 at equivalence point:
     mol Ca(OH)2 = 0.0360L Ca(OH)2 (1.0mol Ca(OH)2 /1L Ca(OH)2)
                            = 0.0360 mol
From the equation for the acid base reaction written as
     2H3PO4 + 3Ca(OH)2 → Ca3(PO4)2 + 6H2O
we use the mole ratio of ammonia and nitrogen dioxide from their coefficients, which is 2 moles of phosphoric acid is to react with 3 moles of calcium hydroxide, to compute for the number of moles of H3PO4:
     mol H3PO4 = 0.0360mol Ca(OH)2 (2mol H3PO4/3mol Ca(OH)2)
                         = 0.0240 mol
Dividing the number of moles of H3PO4 by the number of liters, the molar concentration of H3PO4 is therefore
     [H3PO4] = 0.0240mol / 0.011L
                     = 2.18 M

The correct answer should be "electrons orbit the nucleus in circular motions exactly as shown on the classic bohr model."