The concentration of the sodium hydroxide will be 0.016 M
Stoichiometric problem
First, the concentration of the diluted nitric acid needs to be found.
m1 = 2.0 M, v1 = 100 mL, v2 = 300.0 mL
m2 = 2 x 100/300 = 0.6667 M
The equation of the reaction goes thus: [tex]HNO_3 + NaOH = NaNO_3 + H_2O[/tex]
The mole ratio is 1:1.
Mole of 0.6667 M, 7.05 mL HNO3 = 0.6667 x 0.00705 = 0.0047 mol
Equivalent mole of NaOH = 0.0047 mol
Molarity of 0.0047 mol, 30.0 mL NaOH = 0.0047/0.3 = 0.016 M
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