6B.9 (a)
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6B.9 (a)
For this problem I keep taking the -log of [H3O+] to find the pH and getting -0.176 but the answer book says it's +0.176, what am I doing wrong?
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Re: 6B.9 (a)
I was also really confused on how to get the answer to parts (i) and (ii) of this question. How come you can't use the given value of [H30+]=1.5 mol.L-1 and then substitute into [H30+][OH-]=10^-14 to find [OH-]? I tried doing that, but I kept getting the wrong answer.
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Re: 6B.9 (a)
I think sometimes the concentrations can produce negative numbers when calculating pH or pOH because I don't see any other way to find the pH without getting a negative concentration.
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