Since the acid given, formic acid, has an acid-ionization constant, Ka, we can conclude formic acid is a weak acid that only partially dissociates or does not ionize completely.

For weak acid problems, formic acid’s balanced chemical equation showing the dissociation of the acid into its ions, the acidic proton and its conjugate base, should be written alongside the Ka expression.

**HCOOH**** <—> H+ + HCOO-**

*Ka = [H+][HCOO-]/*

*[HCOOH]*

Equilibrium concentration values in units of moles per liter are used in Ka expressions.

** [H+][HCOO-]**/

*= 1.8×10^-4*

*[HCOOH]** (M H+)(M HCOO-)/ = 1.8×10^-4 (M HCOOH) *

The pH value for the formic acid solution is given. After a weak acid solution reaches equilibrium, the final [H+] concentration is used to calculate the pH value of the solution. Since the problem has given us the final pH value of the solution after equilibration, the equilibrium value for the concentration of hydrogen ions can be calculated and we will work backwards to fill in the rest of the ICE table:

**Equilibrium [H+]**

*[H+]=10^-pH *

* [H+]=10^-2.70*

* [H+]= 2.0×10^-3 M*

HCOOH | <—> | H+ | + | HCOO- | |

i | |||||

c | |||||

e | 2.0E-3 |

**Change in Concentration**

*initial [H+]= 0 *

* equilibrium [H+]= 2.0×10^-3 M*

*change in concentration, x = equilibrium [H+] – initial [H+] = 2.0×10^-3 M – 0 M = 2.0×10^-3 M*

* (2.0E-3)^2/(initial – 2.0E-3) = 1.8×10^-4 *

HCOOH | <—> | H+ | + | HCOO- | |

i | ???? | 0 | 0 | ||

c | -x | +x | +x | ||

e | 2.0E-3 |

HCOOH | <—> | H+ | + | HCOO- | |

i | ???? | 0 | 0 | ||

c | –2.0E-3 | +2.0E-3 | +2.0E-3 | ||

e | initial – 2.0E-3 | 2.0E-3 | 2.0E-3 |

*(initial – 2.0E-3)(1.8E-4) = (2.0E-3)^2*

*initial – 2.0E-3 = 0.0221*

*initial [HCOOH] = 0.0241 M*