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Preface: Buffer solution (acid-base buffer). I am provided with a weak base, which I will designate B. pKa for B's conjugate acid--which I will designate BH--is 8.1, and its mole weight (sic) is 121.1. I'm assuming the latter is the molar mass, though I don't know how that helps me solve this problem.

Problem: 0.1 M (moles/L) of B was dissolved in 1.0 L water, then titrated to pH 8.1 and 7.5. I am to assume no change in volume (sounds like an odd assumption to me, how do you titrate something without [significantly] altering the volume? I always assumed titration is only done by adding a liquid to another liquid, is this not correct?).

I am to find the the concentration of BH at pHs 8.1 and 7.5.

This question looked really easy to me at the first glance, but after a few hours I realized I had no idea on how to go about this.

I've tried calculating the equilibrium concentrations and going from there. This just ends up in x != x in my calculations (where x is the concentration of some substance), which obviously doesn't hold.

I've used the acid dissociation-, base dissociation- and Henderson–Hasselbalch equations, but nothing seems to do the trick. I might not be the brightest person to have grazed the surface of the planet, but I'm confident that I've got a firm grasp on elementary arithmetics.

The answer is provided, and should be 0.05 M at pH = 8.1.

I would highly appreciate somebody showing or explaining the thought process to me.

I tagged this as homework due to the nature of this question, but it isn't. I don't know anybody who knows much chemistry in "real life", so I figured I'd pin my hopes on a friendly soul answering this question. Thanks in advance.

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To answer your question about titrating without adding volume—you don’t. This problem is strange in many ways, but I will attempt to help you to understand.

Since we are working with dissolved $\ce{B}$, I feel as though it makes the most sense to express the reaction with the given by the equation:

$$\ce{B + H2O<=>BH +OH-}$$

So the equilibrium expression of the reaction is given by the equation:

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

If the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid is 8.1, then by mathematical manipulation:

$$\mathrm{p}K_\mathrm{a}=-\log K_\mathrm{a}$$

$$-\log K_\mathrm{a}=8.1$$

$$\log K_\mathrm{a}=-8.1$$

$$K_\mathrm{a}=10^{-8.1}\approx7.94\times10^{-9}$$

Since $\mathrm{p}K_\mathrm{w}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, the $\mathrm{p}K_\mathrm{b}$ of the base should be, and since $\mathrm{p}K_\mathrm{w}=1\times10^{-14}$:

$$1\times10^{-14}=7.94\times10^{-9}\cdot K_\mathrm{b}$$

$$K_\mathrm{b}=\frac{1\times10^{-14}}{7.94\times10^{-9}}\approx1.26\times10^{-6}$$

For the same reason that $1\times10^{-14}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, $\mathrm{14=pH+pOH}$. Given that the pH of the water is 8.1 for our first calculation:

$$14=8.1+\mathrm{p}\ce{OH}$$

$$\mathrm{pOH}=5.9$$ and $$-\log{[\ce{OH-}]}=5.9$$

$$\ce{[OH^{-}]}=\mathrm{10^{-5.9}}\approx1.26\times10^{-6}$$

Since we know that any $\ce{BH}$ that is made must come from the reaction of 1 molecule of $\ce{B}$, then:

$$[\ce{B}]=0.1-x~~~~[\ce{BH}]=x$$

Using the knowledge we have gained from the problem, we can now solve (finally!):

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

$$1.26\times10^{-6}=\frac{[x]\cdot[1.26\times10^{-6}]}{[0.1-x]}$$

$$x=0.1-x$$

$$2x=0.1$$

$$x=0.05$$

So:

$$[\ce{B}]=0.1-0.05=0.05~\mathrm{M}$$

Try using this process to solve for the $\mathrm{pH}$ of 7.5.

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  • $\begingroup$ Thank you very much for your response, it was very helpful to me. I tried a myriad of approaches, including, from the Henderson–Hasselbalch equation, $pH = pK_a + log_{10} \frac{[Base]}{[Acid]} => 8.1 = 8.1 + log_{10}(\frac {0.1}{x}) => x = 0.1$, and I also tried calculating the equilibrium concentrations of acid and base to get $8.1 = 8.1 + log_{10}(\frac {a-x}{b+x})$, where a and b are the equilibrium concentrations before adding the acid. Do we just ignore the equilibrium concentrations because they are relatively small as compared to after having added the acid? $\endgroup$ – HRMD Apr 18 '15 at 3:01
  • $\begingroup$ Well, the Henderson-Hasselbalch equation would indeed work for this problem, and all things considered, this was probably what you should have used. Your problem in your first equation was assuming that at equilibrium, $[Base]=0.1$. You had a better idea with the second equation second equation the $\log{\frac{[Base]}{[Acid]}}$, but you should have used $[Base]=0.1-x$ and $[Acid]=x$. This would give you the same result as in my method. $\endgroup$ – ringo Apr 18 '15 at 3:24
  • $\begingroup$ I have a question related to a subproblem of this problem. If it isn't too much to ask, would you please check out my work and see where my error lies? Problem: Add 0.01 mol of base to the solution. Assume that the pH before this addition was 8.1 (i.e. [B] = [BH] = 0.05 M). I assumed that when adding base, the equilibrium would be displaced, causing more BH to form. From the base dissociation equation I get: $$1.26\times10^{-6} = \frac {(0.05+x)(1.26\times10^{-6}+x)}{0.06-x}$$ As far as I can tell, this should be correct, but when solving the quadratic equation (see next post), $\endgroup$ – HRMD Apr 18 '15 at 14:31
  • $\begingroup$ adding the solution to the existing $OH^-$ concentration, and calculating the pH from there, my answer is off by about .2 pH no matter how many digits I include or exclude from my calculations. I also don't understand why I can't just add my solution for x to [BH] and find the pH directly from $-log_{10}(0.05+x)$, but instead have to add x to $[OH^-]$, calculate the pOH and arrive at the pH from there, since BH is basically the equivalent of $H_3O^+$ in this reaction, no? $\endgroup$ – HRMD Apr 18 '15 at 14:40
  • $\begingroup$ Do you mind me asking what you are getting as your pH calculation? $\endgroup$ – ringo Apr 18 '15 at 16:29

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