Question

Strong Acids and [H'] (and pH) The pH of a strong acid is the concentration of the strong acid. 1.OM HCI makes 1.OM H. Except H2SO4. Its pH is a little lower than that because: H2SO4(aq) H+ (aq) + HS07 (aq) Strong acids generally have K.>>1. We assume K = . The HSO, is a weak acid: HSO2 (aq) + H+ (aq) + S0 - Ka = 1.2 * 10-2 Weak Acids and [H+] (and pH) These are weak electrolytes. What is the pH of 2.OM hypochlorous acid (HOCI)? K. (the acid equilibrium constant) =3.5*108? Part 1. Write the equilibrium expression Part 2. Make the ICE table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. Part 3. Solve for x. Try the small x first! It will work. Part 4. Convert [H+] to pH The pH is: Strong Bases and [OH-] (and pH) The pOH of a strong base is proportional to the concentration of OH made by the dissociation. _M(OH). 1.0*10%M Mg(OH)2 makes 1.0*10PM (Mg2+) and What is the pOH of this solution? Strong bases generally have Ko>>1. We assume Kb= c. What is the pH of this solution? Weak Bases and [OH-] (and pH) These are weak electrolytes. What is the pH of 1.0M methyl amine (H2NCH3)? K. (the base equilibrium constant) =4.38*104? You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. You need to make an ice table. ICE Table: x=[OH ]= pOH= pH=

Answer 1

1 Hore ② H+ + oeo initially am at eam 2(1-x) zie 2x Here, x = degree of dissociation .: Applying law of mass action de ka = [ht] [oce) [Hoce] » Ka = 2xxon 2 (1-) Hoce is very Ka is very small means weak aid. .. (1-10) Thus, Ka = 2x² -

→ x =V1.75 X 10 % = 1.32 x1024 Thues; .(H+] =:2x . . - 2.* 1.32x10* 2.64 x1024 .: pH = -log[ht] = -log (2.64x109) :) = -log (2:54). - log 107 = -0.422 + 4 x logio = -0.422 + 4 x1 pH = 3:58 Ans © Since: Mg(OH), is strong base. So, it would dissociate completely, Mg(OH)2 + Mg²+ + 2010 103 103M 2x103 m

Thus, [OH-] = 2 x 20-3. NA :: pon = log [ón I E -log (2 x103) = -loge - log103 = -0:301-6-3) logio = -0.301 +3X1 IPOH = 2.7] Since pH + pol = 14 i 2 pH = 14 - POH - pH = 14 - 227 → (pH = 11.3] Av. - initially at eam CH₃NH₂ + H₂O ² CH₂ H3 + oHo.... IM (excess) o o 1-24 . x je as degree of dissociation IAS 3

This, ki - [CH3 NH3] [OH-] [CH3NH2] Kb = b = -dxx (1-2) since (-2) i ( weak base) . Thues 1.= = 14:38 x 10-4 - 2.09 x102 . Thus, [OH-] = 2.09x10-2 .. DOH = -log(2.09 x102) = -log (2.09) - log102 =-0:32 +2 . I pOH = 1.68 : 14 - (-68 = 12.32 pH = 14- poh = (pl. = 12:32 A