# Buffer Calculations

AP Chemistry Investigation #15: Household Products' Buffering Activity.

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Dr. Hayek

## Key Questions

• A buffer solution will maintain a constant pH with the addition of small amounts of acid and alkali.

There is a limit to the amount of acid or alkali that the buffer can handle.

The buffer capacity is given by the expression:

$\frac{\mathrm{dn}}{d \left(p H\right)}$

$\mathrm{dn}$ is a tiny amount of added base which results in tiny change in pH shown as $d \left(p H\right)$

If you have a weak acid whose dissociation constant is ${K}_{a}$ then the buffer capacity is given by :

(From Wiki)

${C}_{A}$ is the concentration of the acid.

• Buffers solutions are composed either of a weak acid and its conjugate base, or a weak base and its conjugate acid.

A buffer's capacity to absorb the addition of an acid or a base is proportional to the concentrations of its weak acid - conjugate base components. These components must be in equal or approximatly equal concentrations in the buffer.

Now, the general form for a buffer containing a weak acid and its conjugate base is

$H {A}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{-}$

The Henderson -Hasselbach equation can be expressed as

$\left[p H\right] \left(h \texttt{p} : / s o c r a t i c . \mathmr{and} \frac{g}{c} h e m i s t r \frac{y}{a} c i \mathrm{ds} - \mathmr{and} - b a s e \frac{s}{t} h e - p h - c o n c e p t\right) = p {K}_{a} + \log \left(\frac{\left[c o n j u g a t e . b a s e\right]}{\left[a c i d\right]}\right)$, or

$p H = p {K}_{a} + \log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$, where

${K}_{a}$ - the acid dissociation constant.

So, let's say we need to calculate the pH of a buffer solution that has $0.15 M$ $H F$ (a weak acid) and $0.15 M$ ${F}^{-}$ (its conjugate base). The ${K}_{a}$ for $H F$ is equal to $6.8 \cdot {10}^{- 4}$.

You could do this either by using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart), or by using the Henderson-Hasselbach equation.

$p H = p {K}_{a} + \log \left(\frac{\left[{F}^{-}\right]}{\left[H F\right]}\right) = - \log \left(6.8 \cdot {10}^{-} 4\right) + \log \left(\frac{0.15}{0.15}\right) = 3.17 + \log 1 = 3.17 + 0 = 3.17$

The Henderson-Hasselbach equation is a very useful tool when trying to design a buffer solution that has a certain pH value. Let's assume we want to design a buffer that has a pH of 4.60.

Assuming we can choose any weak acid-conjugate base pair, we'll take acetic acid, $C {H}_{3} C O O H$, and the salt of its conjugate bse, $C {H}_{3} C O O N a$ (its conjguate base being $C {H}_{3} C O {O}^{-}$). We know from acid strenght tables that acetic acid's $p {K}_{a}$ value is 4.75. So,

$p H = p {K}_{a} + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)$

$4.60 = 4.75 + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)$

$\log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right) = 4.60 - 4.75 = - 0.15 \to$

$\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]} = 0.709$

This means that the concentration of the weak acid must be 1.41 times greater than that of the conjugate base in this buffer.

• Buffer solutions are used to calibrate pH meters because they resist changes in pH.

When you use a pH meter to measure pH, you want to be sure that if the meter says pH = 7.00, the pH really is 7.00.

So you use solutions of known pH and adjust the meter to display those values.

Buffers are ideal for this purpose because:

• They are easily prepared for a given pH.
• They are stable for long periods of time.
• They resist changes in pH if you accidentally add a little acid or base or even water.

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