How to calculate the concentration of conjugate acid from given pH and pKa values? Preface: Buffer solution (acid-base buffer). I am provided with a weak base, which I will designate B. $\mathrm{p}K_\mathrm{a}$ for $\ce{B}$'s conjugate acid, which I will designate $\ce{BH}$, 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: $\pu{0.1 M}$ ($\pu{mol/L}$) of $\ce{B}$ was dissolved in $\pu{1.0 L}$ water, then titrated to $\mathrm{pH}$ $8.1$ and $7.5$. I am to assume no change in volume.


The last statement 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 concentration of $\ce{BH}$ at $\mathrm{pH}$ $8.1$ 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 $x \neq x$ 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 arithmetic.

The answer is provided, and should be $\pu{0.05 M}$ at $\mathrm{pH} = 8.1$.

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

$$\ce{B + H2O&lt;=&gt;BH +OH-}$$

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.