Question

Nitric acid (HNO3) is a strong acid that completely ionizes when dissolved in water (pKa = -1.30). Calculate the pH of a solution containing 10 mg/L of HNO3.

Answer 1

1st find the concentration of HNO3

Molar mass of HNO3,

MM = 1*MM(H) + 1*MM(N) + 3*MM(O)

= 1*1.008 + 1*14.01 + 3*16.0

= 63.018 g/mol

mass(HNO3)= 10 mg

= 0.01 g

volume , V = 1 L

number of mol of HNO3,

n = mass of HNO3/molar mass of HNO3

=(0.01 g)/(63.018 g/mol)

= 1.587*10^-4 mol

Molarity,

M = number of mol / volume in L

= 1.587*10^-4/1

= 1.587*10^-4 M

[HNO3 ] = 1.587*10^-4 M

So,

[H+] = 1.587*10^-4 M

use:

pH = -log [H+]

= -log (1.587*10^-4)

= 3.80

Answer: 3.80