Answer:If we have [H+][OH-]= Kw = 1.0 x 10^-14
Then [H+]= Kw/ [OH-]= 1.0x 10^-14/ 1 x 10^-11 =1 x 10^-3 mol/L
And here is the solution - as you can see it is an acidic one :
pH = - log [H+]= - log 1 x 10^-3 = 3 < 7
Step-by-step explanation:
Answer:
[H+] = 0.01mol/L
p[OH] = -log[OH-]
[OH-] = 1*10^-12
pOH = -Log[1*10^-12]
pOH = 12.
But pH + pOH = 14
pH = 14 - pOH
pH = 14 - 12
pH = 2
pH = -Log[H+]
2 = -Log[H+]
Take the anti log of both sides
10⁻² = [H+]
[H+] = 0.01
The hydrogen ion concentration is 0.01mol/L
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