Respuesta :

Answer : The volume of [tex]HNO_3[/tex] needed are, 0.013 L

Explanation :

According to the neutralization law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] = 6.0 M = 6.0 mole/L

[tex]M_2[/tex] = molarity of [tex]KOH[/tex] = 2.0 M = 2.0 mole/L

[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] = ?

[tex]V_2[/tex] = volume of [tex]KOH[/tex] = 39 ml = 0.039 L

conversion : 1 L = 1000 ml

Now put all the given values in the above formula, we get the volume of [tex]HNO_3[/tex]

[tex]6.0mole/L\times V_1=2.0mole/L\times 0.039L[/tex]

[tex]V_1=0.013L[/tex]

Therefore, the volume of [tex]HNO_3[/tex] needed are, 0.013 L

samueladesida43

The volume of HNO3 needed to neutralize 39 mL of 2.0 M KOH is 13mL.

HOW TO CALCULATE VOLUME:

The volume of a substance can be calculated by using the following expression:

CaVa = CbVb

Where:

  1. Ca = concentration of acid
  2. Cb = concentration of base
  3. Va = Volume of acid
  4. Vb = Volume of base

According to this question;

  • Ca = 6M
  • Cb = 2M
  • Va = ?
  • Vb = 39mL

6 × Va = 2 × 39

6Va = 78

Va = 78/6

Va = 13mL.

Therefore, the volume of HNO3 needed to neutralize 39 mL of 2.0 M KOH is 13mL.

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