Question

Assuming complete dissociation, what is the pH of a 4.56 mg/L Ba(OH)2 solution?

Answer 1

Step 1:

1st find the concentration of Ba(OH)2 assuming volume is 1 L

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)= 4.56 mg

= 0.00456 g

number of mol of Ba(OH)2,

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

=(0.00456 g)/(171.316 g/mol)

= 2.662*10^-5 mol

volume , V = 1 L

Molarity,

M = number of mol / volume in L

= 2.662*10^-5/1

= 2.662*10^-5 M

step 2:

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

= 2*2.662*10^-5 M

= 5.324*10^-5 M

step 3:

use:

pOH = -log [OH-]

= -log (5.324*10^-5)

= 4.27

use:

PH = 14 - pOH

= 14 - 4.27

= 9.73

Answer: 9.73