A monoprotic weak acid ( HA ) with a K a of 0.000437 is dissolved in water. The resulting solution has a pH of 2.790. What is the formal concentration ( F ) of HA in solution?

[HA] = 0.0059 M

Explanation:

As we know that we have a weak acid HA, we can use the general equation, as we have Ka:

HA + H₂O > H₃O⁺ + A⁻    Ka = 0.000437

In general terms, this kind of questions are answered using a ICE chart. This usually works when we want to know the final concentration of the acid or the hydronium. However, this actually works if we want the innitial concentration, in this case, we will call this concentration F. Now, let's do first the ICE chart:

       HA + H₂O > H₃O⁺ + A⁻    Ka = 0.000437

i)        F                          0         0

c)      -x                         +x        +x

e)     F-x                         x          x

With this, we can write an expression for the equilibrium constant:

Ka = [H₃O⁺] [A⁻] / [HA]

Replacing the values we have:

0.000437 = x² / F - x

0.000437 = x²/F

Ka is a very small value, so we can neglect the substration of F-x and just put F. This is because we are assuming the value of x will be small too.

So, the values of "x" can be calculated because we have the pH, and the pH is relationed to the H₃O⁺ concentration so:

[H₃O⁺] = antlog(-pH)

[H₃O⁺] = antlog(-2.79) = 0.0016 M

We have the concentration of the hydronium, so we can replace this value in the expression of the equilibrium to solve for F:

0.000437 = (0.0016)² / F

F = 0.0016² / 0.000437

F = [HA] = 0.0059 M

the answer for sodium is 0.186

the answer for rubidium is 0.251