how many liters of 0.3m h3po4 are needed to neutralize 3.5l of 3m naoh

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sebassandin sebassandin · Answer 1 · 2020-06-16T02:38:13+02:00

Answer:

[tex]V_{acid}=11.7L[/tex]

Explanation:

Hello,

In this case, given the reaction:

[tex]3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O[/tex]

We notice a 3:1 molar ratio between sodium hydroxide and phosphoric acid, therefore, at the equivalence point we have:

[tex]n_{base}=3*n_{acid}[/tex]

That in terms of molarity is:

[tex]M_{base}V_{base}=3*M_{acid}V_{acid}[/tex]

Se we solve for the volume of acid:

[tex]V_{acid}=\frac{M_{base}V_{base}}{3*M_{acid}} =\frac{3M*3.5L}{3*0.3M}\\ \\V_{acid}=11.7L[/tex]

Best regards.

dsdrajlin dsdrajlin · Answer 2 · 2020-06-16T02:42:32+02:00

Answer:

11.7 L

Explanation:

How many liters of 0.3M H₃PO₄ are needed to neutralize 3.5L of 3M NaOH?

Step 1: Write the balanced equation

H₃PO₄ + 3 NaOH = Na₃PO₄ + 3 H₂O

Step 2: Calculate the reacting moles of sodium hydroxide

3.5 L of 3 M NaOH were employed. The reacting moles of NaOH are:

[tex]3.5L \times \frac{3mol}{L} = 10.5 mol[/tex]

Step 3: Calculate the reacting moles of phosphoric acid

The molar ratio of H₃PO₄ to NaOH is 1:3. The reacting moles of H₃PO₄ are (1/3) × 10.5 mol = 3.5 mol

Step 4: Calculate the required liters of phosphoric acid

3.5 moles of 0.3 M H₃PO₄ were employed. The required volume of H₃PO₄ is:

[tex]3.5mol \times \frac{1L}{0.3mol} =11.7 L[/tex]