## Problem 14-79 |

To find the K_{a} of trichloroacetic acid:

HCCl_{3}CO_{2} | ⇌ | H^{+} | + | CCl_{3}CO_{2}^{−} | |

I | 0.050 M | 0 | 0 | ||

C | 0.050 M - x | +x | +x | ||

E | 0.050 M - x | x | x |

K_{a}= |
x^{2} | |

0.050 - x |

HClO_{4} is a strong acid, therefore it is completely dissociated in water

0.040 M HClO_{4} = 0.040 M [H^{+}] in solution

Since both HCCl_{3}CO_{2} and HClO_{4} acid have the same pH, then [H^{+}] are the same

Knowing that at equilibrium x = [H^{+}] which is 0.040 M, then:

K_{a}= |
(0.040)^{2} | = 0.16 |

0.050 - 0.040 |