Question

In: Chemistry

Assuming complete dissociation, what is the pH of a 4.21 mg/L Ba(OH)2 solution? pH=

Assuming complete dissociation, what is the pH of a 4.21 mg/L Ba(OH)2 solution?

pH=

Expert Solution

Molar mass of Ba(OH)2 = 171.34 g/mol

Lets assume 1 L of solution.

Mass = 4.21 mg = 4.21*10^-3 g

Number of moles =mass/ molar mass

= (4.21*10^-3)/ 171.34

= 2.457*10^-5 mol

[Ba(OH)2] = number of moles / volume

= 2.457*10^-5 / 1

= 2.457*10^-5 M

Ba(OH)2 ----> Ba2+ + 2 OH-

So,

[OH-]= 2*[Ba(OH)2] = 2*2.457*10^-5= 4.91*10^-5 M

pOH = - log [OH=]

= -log (4.91*10^-5)

= 4.31

pH= 14- pOH

= 14- 4.31

=9.69

Answer: 9.69

queen_honey_blossom answered 1 year ago

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