Question

An aqueous solution at 25 degrees celsius is 12.0% HNO3 by mass and has a density of 1.055 g/mL. What is the pH?

Answer 1

Let volume of solution be 1 L

volume , V = 1 L = 1*10^3 mL

density, d = 1.055 g/mL

we have below equation to be used:

mass = density * volume

= 1.055 g/mL *1*10^3 mL

= 1055.0 g

This is mass of solution

mass of HNO3 = 12.0 % of mass of solution

= 12.0*1055.0/100

= 126.6 g

Molar mass of HNO3 = 1*MM(H) + 1*MM(N) + 3*MM(O)

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

= 63.018 g/mol

mass of HNO3 = 126.6 g

we have below equation to be used:

number of mol of HNO3,

n = mass of HNO3/molar mass of HNO3

=(126.6 g)/(63.018 g/mol)

= 2.009 mol

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 2.009/1

= 2.009 M

So,

[H+] = [HNO3] = 2.009 M

we have below equation to be used:

pH = -log [H+]

= -log (2.009)

= -0.303

Answer: -0.303