Question

43. An aqueous solution of methylamine (CH3NH2) has a pH of 10.65. How many grams of methylamine are there in 100.0 mL of the solution? Calculate the pH and percent ionization of a 0.990 M HNO2 solution. pH = %

Answer 1

PH = 10.85 POH = 14-10.65 POH = 3.35 therefore [OH] = Aut(-3.35)[OH] = 4.467*10^-4 md/L therefore melarity = wt/ MW * V(dm^3) 4467 * 10^-4 = wt/33*0.1 therafore wt = 1.474*10^-3 g therefore HNO_2 is a weat acid with Ka = 4.3*10^-4 therefore HNO_2 +H_2O implies H_3O^+ + NO_2^_ therefore K_a = {H_3O^+]{NO_2^_]/[HNO_2] Initial 0.990 therfore Ka = c alpha * c alpha/0.990(1-alpha) = 0.990)^2 alpha^2 /0.990(1-alpha) therefore alpha < < < 1, there fore 1- alpha =1 therfore ka = 0.990alpha therefore alpha = (4.3*10^-4/0.990)^3/2 = 0.0208 therefore [H_3O^+] = c alpha = 0.02059 therefore P^H = -log[H_3O^+] = 1.686 therfore % ionization = 0.0208/ 0.990 *100 = 2.1%