Could anyone tell me how I would go about calculating the pH of 100.0 mL of 0.100 M $\ce{H2NNH2}$ ($K_\text{b} = 3.0 \times 10^{-6}$).

Furthermore, how would this pH change when 20.0 mL of 0.200 M $\ce{HNO3}$ is added to the solution?

I’m not looking for only an answer, but rather the general method to solve these types of problems

Just to include what I’ve tried:

I first thought that (for the initial solution) $\mathrm{pH} = 14 - \log(\sqrt{K_\text{b} [\ce{H2NNH2}]})$, but that doesn’t seem to be the case.

  • $\begingroup$ That's what I got too (granted, I'm fairly rusty), why do you say that's incorrect? $\endgroup$
    – Kevin
    Commented Oct 10, 2012 at 3:13

1 Answer 1


Concerning the first part of your problem: In the general case, which you requested, we start with the mass action law:

$$ K_\text{b} = \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]} = \frac{[\ce{OH-}]^2}{[\ce{B}]} $$

That’s essentially what you just reformulated to obtain your expression for $\mathrm{pH}$ above. At this point, you made an error in the sign of the second term on the right-hand side – it should be added, not subtracted, i.e. $\mathrm{pH} = 14 + \frac{1}{2} \lg \left( [\ce{B}]K_\text{b} \right)$.

Now comes the important thing: The mass action law uses the concentrations of our components in the equilibrium! What is given as 0.1 M is, however, the starting concentration $[\ce{B}]_0$ before equilibrium, which will of course decrease; $[\ce{HB+}]$ and $[\ce{OH-}]$ will increase by the same amount, which we will denote $x$. Thus, we rewrite:

$$ K_\text{b} = \frac{[\ce{OH-}]^2}{[\ce{B}]} = \frac{x^2}{[\ce{B}]_0 - x} $$

Now, you just solve for $x$, which is the concentration $[\ce{OH-}]$.

One more thing. You will see that $x$ is actually pretty small in this case, so $[\ce{B}]_0 - x \approx [\ce{B}]_0$. By using $K_\text{b} = \frac{x^2}{[\ce{B}]_0}$, we essentially obtain the same result. You can’t always rely on this, however, and if you have stronger acids or bases, this will get important.


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