Calculate the PH of a 0.032 M solution of Ca(OH)2. A) 1.19 B) 12.51 C) 12.81 D)

Question

Calculate the PH of a 0.032 M solution of Ca(OH)2.
A) 1.19 B) 12.51 C) 12.81 D) 1.49 E) none of these
Answer is c but why?
I did -log(0.032) = 1.49 Then 14 - 1.49 = 12.505 which give me B
Calculate the PH of a 0.032 M solution of Ca(OH)2.
A) 1.19 B) 12.51 C) 12.81 D) 1.49 E) none of these
Answer is c but why?
I did -log(0.032) = 1.49 Then 14 - 1.49 = 12.505 which give me B

A) 1.19 B) 12.51 C) 12.81 D) 1.49 E) none of these
Answer is c but why?
I did -log(0.032) = 1.49 Then 14 - 1.49 = 12.505 which give me B

Explanation / Answer

Ca(OH)2(aq) <==> Ca+(aq) + 2OH-(aq)

1 mole of Ca(OH)2 will dissociate 2 moles of OH-

Therefore 0.032 M of Ca(OH)2 will dissociate 0.032x2 = 0.064 M of OH-

pOH = -log[OH-] = -log(0.064) = 1.19

A

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