Question

A chemist dissolves 540. mg of pure nitric acid in enough water to make up 90.mL of solution. Calculate the pH of the solution.Round your answer to 2 significant decimal places.

Answer 1

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)= 540 mg

= 0.54 g

use:

number of mol of HNO3,

n = mass of HNO3/molar mass of HNO3

=(0.54 g)/(63.02 g/mol)

= 8.569*10^-3 mol

volume , V = 90 mL

= 9*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 8.569*10^-3/9*10^-2

= 9.521*10^-2 M

So,

[H+] = 9.521*10^-2 M

use:

pH = -log [H+]

= -log (9.521*10^-2)

= 1.0213

Answer: 1.02