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HI is strong acid . It is completely ionized.
HI(aq) -----------> H^+ (aq) + I^- (aq)
1.556M 1.556M
[H^+] = [HI] = 1.556M
% Ionization = [H3O^+]*100/[HA]
= 1.556*100/1.556 = 100% >>>>answer
4.) What is the % ionization of HI acid if the initial concentration of hydroiodic acid...
Hydroiodic acid (HI) is a strong acid. Which of the following statements are correct (select ALL that apply) (A) a reaction between equal amounts of hydroiodic acid with a weak base gives an acidic solution (B) a hydroiodic acid solution contains both HI molecules and I- ions (C) the iodide ion (I-) is the conjugate base of HI (D) the pH of a 0.01 M solution of hydroiodic acid is -2
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Data Table 1 Solution A.05M Acetic Acid IA (aq) + 1,0 (1) PH : 2.90 1,0' (aq) + (aq) 15 Molarity - Initial Concentration Change quilibrium Concentration 05-x K = [H,01[A] = [x][x] [HA] [.05-x] % ionization = [H0+1*100 [HA Jinitial = [x] * 100 .05 Data Table 2 Solution B 0.025M Acetic Acid PH: 3.00 IA (aq) + 10 (1) 1:0 (aq) + (aq) 1.025 Tolarity - Initial Concentration Change Equilibrium Concentration 1.025-X K = [x][x] [0.025-x] % ionization...
percent ionization of a weak base pH and Percent lonization of a Weak Base Ammonia, NH, is a weak base with a Ky, value of 1.8 x 105 The degree to which a weak base dissociates is given by the base ionization content K) For the generic weak base, B B(aq) + H2O(1) BH(aq) + OH (!) this constant is given by Pan A KDO What is the pH of a 0.165 Mammonia solution? Express your answer numerically to two...
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