Asked by Jonah
A tank contains exactly 10^(6)liters of pure water. 1.0*10^(-6) moles of HCl (g) is bubbled into this tank. What is the resulting pH?
The answer is 7, but I was only able to deduce that based on the fact that water has a neutral pH, but there's a mathematical way to do it and I'm not sure how to go about it.
The answer is 7, but I was only able to deduce that based on the fact that water has a neutral pH, but there's a mathematical way to do it and I'm not sure how to go about it.
Answers
Answered by
DrBob222
HCl is a strong acid meaning that it dissociates 100%; therefore, 1E-6 mol/1E6L = 1E-12M HCl.
HCl ==> H^ + Cl^-
1E-12...0.....0
-1E-12..1E-12..1E-12
..........HOH ==> H^+ + OH^-
I...............1E-12....0
C..................x......x
E..............1E-12+x....x
(H^+)(OH^-) = Kw = 1E-14
(1E-12+x)(x) = 1E-14
Solve the quadratic. I get all of these numbers on the calculator like
9.99995E-8
pH = -log(H^+) = 7.00000.
With no temperature listed I couldn't look up the correct value for Kw at that particular temperature but I think this will do the trick.
HCl ==> H^ + Cl^-
1E-12...0.....0
-1E-12..1E-12..1E-12
..........HOH ==> H^+ + OH^-
I...............1E-12....0
C..................x......x
E..............1E-12+x....x
(H^+)(OH^-) = Kw = 1E-14
(1E-12+x)(x) = 1E-14
Solve the quadratic. I get all of these numbers on the calculator like
9.99995E-8
pH = -log(H^+) = 7.00000.
With no temperature listed I couldn't look up the correct value for Kw at that particular temperature but I think this will do the trick.
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