Find amount of acid and base required to make buffer
Usually when you are preparing a buffer, you can about the pH and the strength (total concentration). Let's say you want to combine acid $\ce{A}$ and its conjugate base $\ce{B}$ to make a buffer with total concentration $C=[\ce{A}]+[\ce{B}]$ and a set $\text{pH}$. 

The Henderson-Hasselbalch equation provides a relationship between the ratio of the two concentrations to the pH:
$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{[\ce{B}]}{[\ce{A}]}\right)$$

Because $[\ce{B}]=C-[\ce{A}]$, we can write, and solve for $[\ce{A}]$ (and then for $[\ce{B}]$).

$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)$$
$$\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)= \ce{pH}-\ce{p}K_a$$
$$\dfrac{C-[\ce{A}]}{[\ce{A}]}=10^{\ce{pH}-\ce{p}K_a}$$
$$C-[\ce{A}]=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]$$
$$C=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]+[\ce{A}]=[\ce{A}]\left(1+10^{\ce{pH}-\ce{p}K_a}\right)$$
$$[\ce{A}]=\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}$$
$$[\ce{B}]=C-\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}=\dfrac{C\left(10^{\ce{pH}-\ce{p}K_a}\right)}{1+10^{\ce{pH}-\ce{p}K_a}}$$

So... if you have a total concentration $C$ and a $\ce{pH}$ in mind for an acid-base pair with a certain $\ce{p}K_a$, just plug them in.