The mass of nitric acid required to make the given solution is 0.0627 g.
The given parameters:
The hydrogen ion (H⁺) concentration of the nitric acid is calculated as follows;
[tex]H^+ = 10^{-pH}\\\\H^+ = 10^{-2.4}\\\\H^+ = 0.00398[/tex]
The molarity of the nitric acid is calculated as follows;
[tex]= 0.00398 \ H^+ \times \frac{1 \ M \ HNO_3}{1 \ H^+} \\\\= 0.00398 \ M[/tex]
The number of moles of the nitric acid is calculated as follows;
[tex]moles = M\times L\\\\moles = 0.00398\ M \ \times \ \frac{250 \ mL}{1000} \\\\moles = 9.95 \times 10^{-4} \ mol.[/tex]
The molar mass of nitric acid is calculated as;
[tex]HNO_3 = (1) \ + (14) \ + (16 \times 3) = 63 \ g/mol[/tex]
The mass of the nitric acid contained in the calculated number of moles is calculated as;
[tex]mass = moles\ \times \ molar \ mass\\\\mass = 9.95\times 10^{-4} \ mol. \ \times \ 63 \ g/mol\\\\mass = 0.0627 \ g[/tex]
Thus, the mass of nitric acid required to make the given solution is 0.0627 g.
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