A 0.1mol/dm^3 aqueous solution of phosphoric (V) acid, H3PO4, is mixed with a 0.1 mol/dm^3 of aqueous solution of sodium hydroxide.

Which mixture will form the salt Na3PO4?

a) 10cm^3 of H3PO4 with 30cm^3 of NaOH
b) 10cm^3 of H3PO4 with 10cm^3 of NaOH
c) 20cm^3 of H3PO4 with 10cm^3 of NaOH
d) 30cm^3 of H3PO4 with 10cm^3 of NaOH

I've written down the equation, and 1 mol of the acid should mix with 3 mols of the hydroxide to form 1 mol of the salt. But i don't understand, shouldn't all the options form the salt no matter the volume? And, the correct answer should be C, and i don't understand at all!

I may have missed something on reading it but I agree that all should form at least some Na3PO4. And I don't see how (c) can be the correct answer.

The correct answer is (a) based on the first chemical equation below.

The mole ratio of NaOH to H3PO4 determines which salt is produced. Only the first one produces Na3PO4:

H3PO4 + 3NaOH ---> Na3PO4 + 3H2O

H3PO4 + 2NaOH ---> Na2HPO4 + 2H2O

H3PO4 + NaOH ---> NaH2PO4 + H2O

To determine which mixture will form the salt Na3PO4, we need to use stoichiometry to calculate the number of moles of reactants and compare them to the stoichiometric ratio in the balanced equation.

The balanced equation for the reaction between phosphoric acid (H3PO4) and sodium hydroxide (NaOH) is:

H3PO4 + 3NaOH -> Na3PO4 + 3H2O

According to the balanced equation, 1 mole of H3PO4 reacts with 3 moles of NaOH to form 1 mole of Na3PO4.

Now let's calculate the number of moles of H3PO4 and NaOH in each option:

a) 10cm^3 of H3PO4 with 30cm^3 of NaOH:

- The number of moles of H3PO4 = 0.1 mol/dm^3 x (10 cm^3 / 1000 cm^3) = 0.001 mol
- The number of moles of NaOH = 0.1 mol/dm^3 x (30 cm^3 / 1000 cm^3) = 0.003 mol

The ratio of moles of H3PO4 to NaOH is 0.001 mol / 0.003 mol = 1:3, which matches the stoichiometric ratio in the balanced equation. Therefore, Option a) will form the salt Na3PO4.

b) 10cm^3 of H3PO4 with 10cm^3 of NaOH:

- The number of moles of H3PO4 = 0.1 mol/dm^3 x (10 cm^3 / 1000 cm^3) = 0.001 mol
- The number of moles of NaOH = 0.1 mol/dm^3 x (10 cm^3 / 1000 cm^3) = 0.001 mol

The ratio of moles of H3PO4 to NaOH is 0.001 mol / 0.001 mol = 1:1. Although there is an equal amount of moles of H3PO4 and NaOH, the stoichiometric ratio in the balanced equation is 1:3. Therefore, Option b) will not form the salt Na3PO4.

c) 20cm^3 of H3PO4 with 10cm^3 of NaOH:

- The number of moles of H3PO4 = 0.1 mol/dm^3 x (20 cm^3 / 1000 cm^3) = 0.002 mol
- The number of moles of NaOH = 0.1 mol/dm^3 x (10 cm^3 / 1000 cm^3) = 0.001 mol

The ratio of moles of H3PO4 to NaOH is 0.002 mol / 0.001 mol = 2:1. The stoichiometric ratio in the balanced equation is 1:3, so Option c) will not form the salt Na3PO4.

d) 30cm^3 of H3PO4 with 10cm^3 of NaOH:

- The number of moles of H3PO4 = 0.1 mol/dm^3 x (30 cm^3 / 1000 cm^3) = 0.003 mol
- The number of moles of NaOH = 0.1 mol/dm^3 x (10 cm^3 / 1000 cm^3) = 0.001 mol

The ratio of moles of H3PO4 to NaOH is 0.003 mol / 0.001 mol = 3:1. The stoichiometric ratio in the balanced equation is 1:3, so Option d) will not form the salt Na3PO4.

Therefore, the correct answer is a), where 10cm^3 of H3PO4 is mixed with 30cm^3 of NaOH. This option has the correct ratio of reactants to form the salt Na3PO4.