First we must write a balanced chemical equation for this reaction
The mole ratio for the reaction between and
is 1:2. This means 1 moles of
will neutralize 2 moles
. Now we find the moles of each reactant based on the mass and molar mass.
The was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.
When 0.34 of HNO₃ is titrated to equivalence using 0.14 l of 0.1 m NaOH then the concentration of HNO₃ is 0.041 M
The reaction of neutralization of HNO₃ with NaOH is
HNO₃ + NaOH → H₂O + NaNo₃
When 1 mole of HNO₃ react with 1 mole of NaOH, based on chemical rection the moles of NaOH at equivalence point are equal to moles of HNO₃ present in solution: -
With the mole and volumes, we can find molarity as follows:
Moles of NaOH = moles HNO₃
⁼ 0.14 L X (0.1 mol NaOH/L) = 0.014 mole NaOH
=0.014 mol HNO₃
Molarity: -
= 0.041 M
Thus, from above solution we concluded that the concentration of HNO₃ solution is 0.041 M.
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