Decide whether solutions of the ammonium nitrite are acidic, neutral, or basic.Kb for NH3 is 1.8x10^-5. Ka for HNO2 is 4.5x10^-4. acidicneutralbasic Decide whether solutions of the anilinium cyanate are acidic, neutral, or basic. Kb for C6H5NH2 is 4.2x10^-10. Ka for HCN is 4.9x10^-10. acidicneutralbasic

pH is the negative logarithm of the hydrogen ion concentration. It is a measure of acidity or alkalinity in a solution. The acidic solution will have the greater concentration of hydrogen ions and lower the pH. pH ranges from 0 to 14. pH upto 6 is acidic, pH=7 is neutral after 7, the pH is basic.1. Ammonium nitrite is the conjugate acid of a weak base (NH4+) and the conjugate base of a weak acid (NO2-). NH4NO2 → NH4+ + NO2- Kb of NH3 = 1.8×10-5 Ka of HNO2 = 4.5×10-4 The formula for the pH of a salt of a weak acid and a weak base is: pH = 7+12pKa-pKbpKa is the dissociation constant for a weak acidpKb is the dissociation constant for a weak base pKa HNO2 = - log (4.5×10-4) = 3.35pKb NH3 = - log (1.8×10-5) = 4.74 Therefore, pH = 7+12pKa-pKbpH = 7+123.35-4.74 = 6.31 Therefore, solutions of the ammonium nitrite is slightly acidic. 2. Anilinium cyanate is the conjugate acid of a weak base (C6H5NH3+) and the conjugate base of a weak acid (OCN-). C6H5NH3OCN → C6H5NH2 + HOCN Kb of C6H5NH2 = 4.2×10-10 Ka of HOCN = 4.9×10-10 The formula for the pH of a salt of a weak acid and a weak base is: pH = 7+12pKa-pKbpKa is the dissociation constant for a weak acidpKb is the dissociation constant for a weak base pKa HOCN = - log (4.9×10-10) =9.309pKb C6H5NH2 = - log (4.2×10-10) = 9.376 Therefore, pH = 7+12pKa-pKbpH = 7+129.309-9.376 = 7.00 Therefore, solutions of the anilinium cyanate is neutral.