Question

A chemist dissolves 469. mg of pure barium hydroxide in enough water to make up 50. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 o°C.) Round your answer to 2 significant decimal places x10 ?

Answer 1

Molar mass of Ba(OH)2,

MM = 1*MM(Ba) + 2*MM(O) + 2*MM(H)

= 1*137.3 + 2*16.0 + 2*1.008

= 171.316 g/mol

mass(Ba(OH)2)= 469 mg

= 0.469 g

use:

number of mol of Ba(OH)2,

n = mass of Ba(OH)2/molar mass of Ba(OH)2

=(0.469 g)/(1.713*10^2 g/mol)

= 2.738*10^-3 mol

volume , V = 50 mL

= 5*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 2.738*10^-3/5*10^-2

= 5.475*10^-2 M

This is the concentration of Ba(OH)2

So,

[OH-] = 2*[Ba(OH)2]

= 2* 5.475*10^-2 M

= 0.1095 M

use:

pOH = -log [OH-]

= -log (0.1095)

= 0.9606

use:

PH = 14 - pOH

= 14 - 0.9606

= 13.0394

Answer: 13.04