To calculate the acid-ionization constant for cyanic acid, HOCN, the balanced chemical equation showing the ionization of cyanic acid is needed so we can then write the acid-ionization expression to eventually solve for Ka, the acid-ionization constant.

**HOCN (aq) <—> H+ (aq) + OCN- (aq)**

**Ka = [H+][OCN-] / [HOCN]**

To begin to piece together the equilibrium values needed for the Ka expression, an ICE table is used to map out the information given. An initial solution of 1.00 E-2 M HOCN is given:

HOCN | <–> | H+ | OCN- | |

initial | 1.00E-2 M | 0 M | 0 M | |

change | -x | +x | +x | |

equilibrium |

The pH of acid-base reactions is measured after a solution has reached equilibrium so we can use the pH value of 2.77 given to calculate the equilibrium concentration of hydrogen ions:

10^-pH = [H+] therefore, 10^-2.77 = [H+] = 1.698 E-3 M

HOCN | <–> | H+ | OCN- | |

initial | 1.00E-2 M | 0 M | 0 M | |

change | -x | +x | +x | |

equilibrium | 1.698E-3 M |

The initial concentration of H+ is 0 M and since we now know the equilibrium concentration of H+, we can solve for the change variable, x, needed to fill in the rest of the values in the ICE table:

0 + x = 10^-2.77 therefore, x = 10^-2.77 = 1.698E-3 M

HOCN | <–> | H+ | OCN- | |

initial | 1.00E-2 M | 0 M | 0 M | |

change | – 1.698E-3 M | + 1.698E-3 M | + 1.698E-3 M | |

equilibrium | 8.30E-3 M | 1.698E-3 M | 1.698E-3 M |

Now, that the equilibrium values are known for cyanate, OCN-, hydrogen ion, H+, and cyanic acid, HOCN, the acid-ionization equilibrium expression can be used to calculate the value of Ka, the acid-ionization constant.

**HOCN (aq) <—> H+ (aq) + OCN- (aq)**

* [H+] = [OCN-] = 1.698E-3 M *

* [HOCN] = 8.30E-3 M *

**Ka = [H+][OCN-]/ [HOCN]**

*Ka= ( 1.698E-3 M)^2/ (8.30E-3 M)*

*HOCN Ka = 3.47E-4*