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An aqueous solution containing 0.0020 mol/L barium hydroxide, what is the pH? I do the following: 14-(-log(0.0020))=11.3pH. The answer is 11.6.

2.0*10^-3 mol/L sodium hydroxide solution, what is the pH? I do the following: 14-(-log(0.002))=11.3pH. The answer is 9.3

When I solve for pH of solutions with acidic solutes, I get the accurate answer, such as in the question: 2.0*10^3 mol/L nitric acid solution, where I do: -log(0.002)=2.7pH, which matches the answer, 2.7.

My question is, why am I getting the discrepancy between the calculated and the given answer when calculating pH for solutions with base solute? Or am I not doing anything wrong?

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Because barium hydroxide is $\ce{Ba(OH)2}$

The book is assuming each mole of barium hydroxide release 2 moles of hydroxide ions, while you are assuming only 1 mole is released.

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  • $\begingroup$ Please disambiguate. $\endgroup$
    – Arthur Alex Karapetov
    Commented Jun 2, 2016 at 22:21
  • $\begingroup$ @ArthurAlexKarapetov OK, I added to the answer. $\endgroup$
    – DavePhD
    Commented Jun 2, 2016 at 22:40
  • $\begingroup$ I see. But what about sodium hydroxide (NaOH)? There is only 1 hydroxide ion released there. In my question, it doesn't work out either. Please help understand why? $\endgroup$
    – Arthur Alex Karapetov
    Commented Jun 2, 2016 at 22:45
  • $\begingroup$ Your answer is right for 0.002M NaOH $\endgroup$
    – DavePhD
    Commented Jun 2, 2016 at 22:49

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