## When pH is known; solving for K

A pH value is always an *equilibrium* value, so knowing a pH is similar to knowing the equilibirium values in the examples on the previous page.

You may need to account for the dissociation of your acid or base (if you’re only given a starting or *formal* concentration) or convert from $H^+$ to $OH^-$ concentrations (if you are given a pH but your compound is a base).

Determination of *K*_{a} or *K*_{b} from pH

The pH of a 0.0516-*M* solution of nitrous acid, HNO_{2}, is 2.34. What is its *K*_{a}?

Solution

The nitrous acid concentration provided is a *formal* concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as “initial” values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as *approximately* zero because a small concentration of H_{3}O^{+} is present (1 × 10^{−7} *M*) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.

The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an “equilibrium” value for the ICE table:

$$[H_3O^+]=10^{-2.34}=0.0046\;M$$The ICE table for this system is then

Finally, calculate the value of the equilibrium constant using the data in the table:

$$K_a=\frac{[H_3O^+][NO_2^-]}{[HNO_2]}=\frac{(0.0046)(0.0046)}{(0.0470}=4.6×10^{-4}$$### Check Your Learning

The pH of a solution of household ammonia, a 0.950 M solution of NH_{3,} is 11.612. What is *K*_{b} for NH_{3}?

## Answer:

$K_b=1.8×10^{-5}$