calculate the ph at which mg(oh)2 just begin to precipitate from a 0.100m mg(no3) solution by addition of naoh
Answer (1)
28th Dec, 2019
Hello,
The reaction is taking place between Mg(NO3)2 and NaOH
Mg(NO3)2 + 2NaOH Mg(OH)2 + 2(NaNO3)
Since, NaNO3 is soluble in water, it will be in aqueous state in above reaction. So, the only solid material in the product side will be Mg(OH)2
We know that,
[ OH -] = √ Ksp / m ( Mg(OH)2)......... ( where, m is Molarity )
So, [ OH - ] = √ ( 5.61 x 10^-12 ) / 0.1
Hence, [ OH - ] = 7.5 x 10^-6 M
Now, pOH = -log [ OH - ]
So, pOH = 5.2
Also, We know that,
pH + pOH = 14
Hence, pH = 14 - 5.2
So, pH = 8.8
Hence, the pH value is 8.8
Best Wishes.
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