What is the molarity of an aqueous phosphoric acid solution if 13.13 ml is completely neutralized by 19.64 ml of 0.09291 m naoh? collapse question part part 1 incorrect answer. did you write the correct molecular formula for phosphoric acid? write a balanced molecular equation for the reaction. be sure to include the appropriate phases for all reagents and products.

The molarity of this solution is 0.0695 M

Explanation:

Step 1: The balanced equation

H3PO4(aq) + 2NaOH(aq) → Na2HPO4(aq) + 2H2O(l)

Step 2: Given data

Volume of phosphoric acid (H3PO4) = 13.13 mL

volume of NaOH = 19.64 mL

Molarity of NaOH = 0.09291 M

Step 3: Calculate moles of NaOH

Number of moles = Molarity * volume

Number of moles = 0.09291 M * 19.64 mL *10^-3 L

Number of moles = 0.001825 moles

Step 4: Calculate moles of H3PO4

For 2 moles of NaOH there is 1 mole of H3PO4 consumed to produce 1 mole Na2HPO4

So for 0.001825 moles NaOH, there is consumed 0.001825/2 = 0.0009125 moles of H3PO4

Step 5: Calculate molarity of H3PO4

Molarity = number of moles / Volume

Molarity = 0.0009125 moles / (13.13 *10^-3L) = 0.0695 M

The molarity of this solution is 0.0695 M

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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