Buffers control hydrogen ion concentration in solution. pH indicator dyes provide useful models for the effect of buffers. Adding acid or base to an indicator dye changes its color abruptly. In analogous fashion acid or base can denature biomolecules. Adding buffer stabilizes dye color to small additions of acid or base in the same way it stabilizes biomolecules.
pH and salt concentration are important in virology. pH extremes disrupt many viruses. Low salt precipitates some viruses and high salt disrupts others. Extracting viruses from tissue releases vacuolar acids which can precipitate or disrupt viruses. Acid stable viruses, however, are effectively "clarified" by acidification which precipitates proteins and membranes. Bromoviruses are unstable above pH 6.5 and must be stored at lower pHs. Study of bromovirus swelling above pH 7 requires careful control of pH and salt concentration.
Working with solutions requires familiarity with acids, bases and salts. Acids donate hydrogen ions. The most common acids in biology have carboxyl (-COOH), sulfate (-SO4H) and phosphate (-PO4H) groups. Bases accept hydrogen ions. The most common bases have amino (-NH2) groups. Salts exist in solution as ions (other than H+ or OH-).
Hydrogen ion concentrations in aqueous solution vary over more than 10 orders of magnitude and are most conveniently handled as logarithms. The concept of pH was devised by Soren Sorenson in about 1909.
Ionization of weak acids and bases can be described by equilibrium constants. In chemistry, equilibrium constants are expressed as dissociation constants. For example:
HA → A- + H+
Ka = [A-][H+]/[HA] 
When [A-] = [HA], then Ka = [H+]. The pKa of a weak acid is the negative log of its Ka. Since pH = - log [H+] , pKa = -log Ka., when [A-] =[HA], then pH = pKa. On addition of [H+], [A-] or [HA], the equilibrium readjusts. Adding, for example, a little strong acid (H+) to the solution increases the numerator. To satisfy equilibrium, some H+ combines with A-, decreasing [A-] in the numerator and increasing [HA] in the denominator to re-establish equilibrium.
Acetic acid (pKa = 4.73) is a common weak acid. Its ionization is best quantitated by taking the log of both sides of equation :
log Ka = log [H+] + log ([A-]/[HA]) 
Substituting pKa = -log Ka and pH = -log [H+]:
pH = pKa + log ([A-]/[HA]) 
"Buffer problems", i.e. calculating pH of solutions of weak acids or weak bases, are commonly solved with equation , the Henderson Hasselbalch equation.
A simple way to make acetate buffers is to mix sodium (ammonium, potassium, etc.) acetate with acetic acid. Mixing equimolar acid and acetate gives a pH of 4.73 (from equation ). This may seem surprising. How can the pH be independent of the total concentration? Clarify in your mind why this happens. Qualitatively excess acid gives a pH below 4.73, excess acetate gives a pH above 4.73. Fixing the ratio of [acetate]/[acetic acid] fixes the pH.
Table of deviation of pH from pK as a function of [A-]/[HA]
More convenient table
[A-]/[HA] pH [A-]/[HA] pH 1.0 0.0 x 1.0 0.0 1.26 0.1 0.8 -0.1 1.58 0.2 0.63 -0.2 2.0 0.3 0.5 -0.3 2.5 0.4 0.4 -0.4 3.2 0.5 0.32 -0.5 4.0 0.6 0.25 0.6 5.0 0.7 0.2 -0.7 6.3 0.8 0.16 -0.8 7.9 0.9 0.125 -0.9 10 1.0 0.1 -1.0 12.6 1.1 0.08 -1.1 15.8 1.2 0.063 -1.2
You can determine pH from this table if you know the pKa and the proportion of the acid which is ionized.
Sample problem: How could you make an acetate buffer of pH 5 and a salt concentration of 0.1M? Solution: First think through the problem qualitatively. The pH is above the pK, therefore there should be less acetic acid than acetate. Then think about it quantitatively. The required pH is about 0.3 pH units above the pKa. If pH is 0.3 units above the pKa, then from the table [A-]/[HA] should be about 2. To make the solution, begin with 0.1M sodium acetate (i.e. 0.1M salt) and add half this amount of acetic acid (0.05M). This gives a pH of 5.03 which is good enough for most purposes. For a more exact answer solve equation . The answer should be slightly more than 0.05M, i.e. the solution is a little more acid than pH 5.03.
Weakly acidic or weakly basic groups are extremely important in biology. Their ionization state depends on hydrogen ion concentration. For example, the carboxyl group of acetic acid is protonated at low pH and ionized at "high" pH. Since these forms differ in ability to form ionic and hydrogen bonds, the state of weakly acidic and weakly basic groups profoundly affects the properties of biomolecules.
Second sample problem: What is the pH of 0.1M acetic acid? When acetic acid is diluted into water it equilibrates by dissociation. Looking at equation  there seems to be one equation and three unknowns; however, two additional conditions must be fulfilled. Firstly, charges must balance. That is, total negative charges must equal total positive charges. Therefore, [H+] = [A-]. Also mass must be conserved. This means [HA] + [A-] = 0.1M, the amount of acetic acid added to begin with. Substituting both conservation conditions into equation :
pH = pKa + log([H+]/(0.1 - [H+])) [a]
Since the final pH will close to the pKa, the expression(0.1 - [H+]) can be simplified to (0.1).
pH = pKa + log([H+]/0.1) [b]
Since log(a/b) = loga - logb, pH = pKa + log[H+] - log(0.1) [c]
substituting pH = -log[H+],
pH = pKa - pH - log(0.1). [d]
2pH = pKa - log(0.1) [e]
2pH = 4.73 -(-1) = 4.73 + 1 = 5.73 [f]
pH = 2.86
This agrees with the assumption that [H+] is small. To solve [a] more precisely, substitute the value of [H+] into [a] and recalculate. Most situations do not require such precision.
pKas of some common monobasic weak acids