To determine the pH of a solution which has 0.195 M hc2h3o2 and 0.125 M kc2h3o2, we use the ICE table and the acid dissociation constant of hc2h3o2 <span>to determine the concentration of the hydrogen ion present at equilibrium. We do as follows:
HC2H3OO = H+ + </span>C2H3OO-
KC2H3OO = K+ + C2H3OO-
Therefore, the only source of hydrogen ion would be the acid. We use the ICE table,
HC2H3OO H+ C2H3OO-
I 0.195 0 0.125
C -x +x +x
------------------------------------------------------------------
E 0.195-x x 0.125 + x
Ka = <span>1.8*10^-5 = (0.125 + x) (x) / 0.195 -x
x = 2.81x10^-5 M = [H+]
pH = - log [H+]
pH = -log 2.81x10^-5
pH = 4.55
Therefore, the pH of the resulting solution would be 4.55.</span>
Your answer should be <span>sublimation</span>
Answer:
TABLE OF CONJUGATE ACID-BASE PAIRS Acid BaseK a (25oC) HClO 4ClO 4 – H 2 SO 4HSO 4 – HCl Cl– HNO 3NO 3 – H 3 O +H 2 O H 2 CrO 4HCrO 4 –1.8 x 10–1
Explanation:
Gold is a soft metal. Both malleable and ductile
