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3 years ago

14

If the Ksp of NaCl is experimentally determined to be 43.9, then what is the concentration of Na (in M) when it begins to crysta

llize out of solution

1 answer:

serg [7]3 years ago

4 0

Answer:

6.63 M

Explanation:

NaCl(s) ---> Na^+(aq) + Cl^-(aq)

Given that [Na^+] = [Cl^-] = s

Where s= concentration of the both ions

Ksp = s^2

s= √Ksp

s= √43.9

s= 6.63 M

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