1. NaOH = baza tare care disociaza aproape complet in solutie apoasa unde [HO-] = Conc.baza
=> pOH = -lg[OH], pH + pOH = 14
a)
pOH = -lg(0,08) = -(-1,1) = 1,1
=> pH = 14-1,1 = 12,9
b)
Vs = 0,5 L, md = 2g
stim ca Cm = md/miuxVs
= 2/40x0,5 = 0,091 M
=> pOH = -lg(0,091) = -(-1,04) = 1,04
=> pH = 14 - 1,04 = 12,96
2.
a) HI = acid slab
=> stim ca [H+] = [tex]\sqrt{Ca*Ka}[/tex]
pH = -lg[H+] = 2 => [H+] = 10la-2 M
=> [H+]² = Ca*Ka, pKa = -lgKa
=> Ka = 10¹¹
=> Ca = [H+]²/Ka = (10⁻²)²/10¹¹
= 10⁻¹⁵ M
b) HCl = acid tare unde [H3O+] = Ca
din pH = -lg[H3O+] => [H3O+] = 10la-pH
= 10⁻⁰⁵ = 3,16 M
c) pH = 13 => pOH = 14-13 = 1
stim ca [HO-] = Cb = 10la-pOH
= 10⁻¹ M
3.
a.
n moli 0,15 moli
HCl + NaOH --> NaCl + H2O
1 1
=> n = 0,15 moli HCl
Cm = n/Vs => Vs = n/Cm
= 0,15/0,2 = 0,75 L sol. HCl
b.
n moli 2,24 g
HCl + KOH --> KCl + H2O
1 56
=> n = 0,04 moli HCl
Cm = n/Vs => Vs = n/Cm
= 0,04/0,2 = 0,2 L sol. HCl
