Question

What is the pH of a solution in which 313 mL of HCl(g), measured at 27.4 ∘C and 1.02 atm, is dissolved in 1.5 L of aqueous solution?

Express your answer to two decimal places.

Answer 1

Given:

P = 1.02 atm

V = 313.0 mL

= (313.0/1000) L

= 0.313 L

T = 27.4 oC

= (27.4+273) K

= 300.4 K

find number of moles using:

P * V = n*R*T

1.02 atm * 0.313 L = n * 0.08206 atm.L/mol.K * 300.4 K

n = 1.295*10^-2 mol

Now use:

[HCl] = number of mol of HCl / volume in L

= 1.295*10^-2 mol / 1.5 L

= 8.63*10^-3 M

So,

[H+] = 8.63*10^-3

use:

pH = -log [H+]

= -log (8.63*10^-3)

= 2.064

Answer: 2.06