Question

f 23.6 mL of 0.200 M NaOH is required to neutralize 10.00 mL of a H3PO4 solution , what is the concentration of the phosphoric acid solution?Start by balancing the equation for the reaction: H3PO4(aq) + NaOH(aq) → Na3PO4(aq) + H2O(l)

Answer 1

`M_(H_3PO_4)=0.157M`

Step-by-step explanation:

Hello,

In this case, the balanced chemical reaction is:

`H_3PO_4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)`

Therefore, we compute the moles of used NaOH by the 0.2000-M solution:

`n_(NaOH)=0.200(mol)/(L)*0.0236L=4.72x10^(-3)molNaOH`

Then, we compute the moles of neutralized phosphoric acid by their 1:3 molar ratio:

`n_(H_3PO_4)=4.72x10^(-3)molNaOH*(1molH_3PO_4)/(3molNaOH)=1.57x10^(-3)molH_3PO_4`

Finally, the concentration:

`M_(H_3PO_4)=(1.57x10^(-3)molH_3PO_4)/(0.010L)\\ \\M_(H_3PO_4)=0.157M`

Best regards.

Answer 2

Step-by-step explanation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

mole of NaOH = 23.6 * 10 ⁻³L * 0.2M

= 0.00472mole

let x be the no of mole of H3PO4 required of 0.00472mole of NaOH

3 mole of NaOH required ------- 1 mole of H3PO4

0.00472mole of NaOH ----------x

cross multiply

3x = 0.0472

x = 0.00157mole

[H3PO4] = mole of H3PO4 / Vol. of H3PO4

= 0.00157mole / (10*10⁻³l)

= 0.157M

The concentration of unknown phosphoric acid is 0.157M