PH at the equivalence point?

Find the pH of the equivalence point and the volume of .0388 M KOH needed to the the equivalence point in...?

• Answer:

  HNO2 and KOH react in a 1:1 mole ratio: HNO2 + KOH ==> H2O + KNO2 moles HNO2 = M HNO2 x L HNO2 = (0.0390)(0.00234) = 0.0000913 moles HNO2 Since KOH and HNO2 react in a 1:1 mole ratio, then moles HNO2 = moles KOH added. moles KOH = M KOH x L KOH added 0.0000913 = (0.0388)(L KOH added) L KOH added = 0.00235 L = 2.35 mL KOH At the equivalence point, all you have is a solution containing the salt KNO2 (see reaction above). The solution volume = 2.34 mL + 2.35 mL = 4.69 mL. Because the KNO2 was produced by reacting a weak acid (HNO2) with a STRONG base (KOH), the salt solution will be alkaline (basic). That means that NO2- (K+ is just a spectator ion) will react with water to produce HNO2 and OH-. Molarity NO2- = moles NO2- / L of solution = 0.0000913 / 0.00469 = 0.0195 M Molarity . .. . . . . . .NO2- + H2O <==> HNO2 + OH- Initial . . . . . . . . . .0.0195 . . . . . . . . . . . . .0 . . . . . .0 Change . . . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . . .x Equilibrium . . . . .0.0195-x . . . . . . . . . . . . x . . . . . .x Since NO2- is acting as a base (producing OH-), we need Kb for NO2-. Ka HNO2 x Kb NO2- = 1 x 10^-14 Kb = (1 x 10^-14) / (7.1 x 10^-4) = 1.4 x 10^-11 Kb = [HNO2][OH-] / [NO2-] = (x)(x) / (0.0195-x) = 1.4 x 10^-11 Because Kb is so small, we can neglect the -x term to simlify the math. x^2 / 0.0195 = 1.4 x 10^-11 x^2 = 2.7 x 10^-13 x = 5.2 x 10^-7 = [OH-] pOH = -log[OH-] = -log (5.2 x 10^-7) = 6.28 pH = 14.00 - pOH = 14.00 - 6.28 = 7.72 . . .slightly alkaline as we expected.