12.50 mL of 0.100 mol/L NaOH is need to neutralize 50.00 mL of nitric acid solution write a chemical equation for this reaction calculate the molar concentration of the nitric acid solution

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This is a titration problem where acid is allowed to be neutralized by a base. The chemical equation for this reaction is

 HNO3 + NaOH --> H2O + NaNO3

To get the value of the concentration of the nitric acid solution we have to use the amount of NaOH solution...

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This is a titration problem where acid is allowed to be neutralized by a base. The chemical equation for this reaction is

 HNO3 + NaOH --> H2O + NaNO3

To get the value of the concentration of the nitric acid solution we have to use the amount of NaOH solution used in the reaction.

In neutralization reaction:

 

Moles acid = moles base

 

Therefore:

 

Moles NaOH = moles HNO3

 

Remember that, Molarity = moles/volume(L)

So, moles = Molarity * Volume (L)

 

(Molarity x volume)for NaOH = (Molarity x Volume) for HNO3

Substituting the value of the given:

0.100 moles/L*(12.5/1000)L = Molarity*(50/1000)L

Then we solve for the Molarity of HNO3

 

Molarity of HNO3 = (0.100x 0.0125)/(0.05)

Molarity of HNO3 = 0.025 moles/L

 

 

 

 

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