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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)

Respuesta :

shallomisaiah19

Answer:

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

sebassandin

Answer:

[tex]M_{H_3PO_4}=0.157M[/tex]

Explanation:

Hello,

In this case, the balanced chemical reaction is:

[tex]H_3PO_4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)[/tex]

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

[tex]n_{NaOH}=0.200\frac{mol}{L}*0.0236L=4.72x10^{-3}molNaOH[/tex]

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

[tex]n_{H_3PO_4}=4.72x10^{-3}molNaOH*\frac{1molH_3PO_4}{3molNaOH}=1.57x10^{-3}molH_3PO_4[/tex]

Finally, the concentration:

[tex]M_{H_3PO_4}=\frac{1.57x10^{-3}molH_3PO_4}{0.010L}\\ \\M_{H_3PO_4}=0.157M[/tex]

Best regards.

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