Answer:
1.4
Explanation:
Mass of pure nitric acid = 751mg
Volume of solution = 290mL
Unknown:
pH of the solution = ?
Solution:
To solve this problem, we need the concentration of the acid in the aqueous form.
This is given by molarity;
Molarity = [tex]\frac{number of moles }{volume}[/tex]
Since the number of moles of nitric acid is unknown, we can easily solve for it.
Number of moles of nitric acid = [tex]\frac{mass}{molar mass}[/tex]
molar mass of HNO₃ = 1 + 14 + 3(16) = 63g/mol
mass of nitric acid = 751mg = 0.751g
Number of moles = [tex]\frac{0.751}{63}[/tex] = 0.012mole
Volume of solution = 290mL = 0.29dm³
Now molarity of the solution = [tex]\frac{0.012}{0.29}[/tex] = 0.041moldm⁻³
Since:
pH = -log [H₃O⁺]
HNO₃ + H₂O → H₃O⁺ + NO₃⁻
1moldm⁻³ 1moldm⁻³ 1moldm⁻³
0.041moldm⁻³ 0.041moldm⁻³ 0.041moldm⁻³
pH = -log[0.041] = 1.4
