Answer: The volume of [tex]NaOH[/tex] required is 25.0 ml
Explanation:
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity [tex]HNO_3[/tex] = 1
[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] solution = 2.00 M
[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] solution = 50.0 ml
[tex]n_2[/tex] = acidity of [tex]NaOH[/tex] = 1
[tex]M_1[/tex] = molarity of [tex]NaOH[/tex] solution = 4.00 M
[tex]V_1[/tex] = volume of [tex]NaOH[/tex] solution = ?
Putting in the values we get:
[tex]1\times 2.00\times 50.0=1\times 4.00\times V_2[/tex]
[tex]V_2=25.0ml[/tex]
Therefore, volume of [tex]NaOH[/tex] required is 25.0 ml
The volume of 4.00 M NaOH is required to exactly neutralize 50.0 ml of a solution of a 2.00 M solution of [tex]HNO_3[/tex] is 25.0 ml.
[tex]HNO_3[/tex] is nitric acid also known as the spirit of niter. It is colorless in its original form but appears yellow-colored.
By the neutralization law
[tex]n_1M_1V_1 = n_2M_2V_2[/tex]
n1 = 1
M1 = 2.00
M2 = 4.00
V1 = 50.0 ml
n2 = NaOH = 1
Putting the value in the formula
[tex]1 \times 2.00 \times 5.00 ml = 1 \times 4.00 \times V_2\\\\V_2 = 25.0\; ml[/tex]
Thus, the volume of NaOH required is 25.0 ml.
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