Question

If 25.15mL of 0.500M NaOH was required to neautralize 50.00mL of an H3PO4 solution, what is the molarity of the acid solution?

Calculate the % of H3PO4 (w/w) of the phosphorus acid solution from the previous problem

Answer 1

i)

Molarity of H3PO4 = 0.0838M

Explanation

The balanced equation for the reaction between H3PO4 and NaOH is as follows

H3PO4(aq) + 3NaOH(aq) <--------> Na3PO4(aq) + 3H2O(l)

From this balanced reaction we know that 1mole of H3PO4 reacts with 3 moles of NaOH

Now ,calculate the number of moles of H3PO4 present in 50ml H3PO4 solution and number of moles of NaOH consumed

we know that molarity is defined as the number of moles of solute per liter of solution, so

moles of NaOH consumed = (0.500mol/1000ml)×25.15ml = 0.012575mol

Number of moles of H3PO4 present in 50ml solution = No of moles of NaOH consumed/3

Number of moles of H3PO4 present in 50ml solution = 0.012575mol/3 = 0.004192mol

Therefore,

Molarity of H3PO4 = (0.004192mol/50ml)× 1000ml = 0.0838M

ii)

% of H3PO4 (w/w) = 0.815℅

Explanation

In the above problem , we calculated no of moles of H3PO4 present in 50ml solution as 0.004192, now calculate the mass of H3PO4 using the following formula

mass = Number of moles × Molar mass

Molar mass of H3PO4 = 97.998g/mol

mass of H3PO4 present in 50ml solution = 97.998g/mol × 0.004192mol = 0.4108g

Mass of water for 50ml = 50g

Total mass of 50ml H2SO4 solution = mass of water + mass of H3PO3

Total mass of 50ml H2SO4 solution = 50g + 0.4108g = 50.4108g

℅ of H3PO4(w/w) = (mass of H3PO4/total mass)×100

% of H3PO4(w/w) = (0.4108g/50.4108g)×100 = 0.815℅