6. Zuren en basen

6. Zuren en basen 6. Zuren en basen

10.09.2013 Views

Buffer + sterk zuur of base • Kleine pH-wijziging door grote [Ac - ] e en [HAc] e Toevoeging [H 3O + ] ⇒ evenwicht naar rechts ⇒ [H 3O + ] daalt opnieuw (tot Ac - opgebruikt) • [Ac - ] 0 = 0,1 mol/l en [HAc] 0 = 0,1 mol/l V = 50 ml ⇒ n Ac- en n HAc + 1 ml [HCl]=1 mol/l ⇒ n H3O+ ⇒ evenwicht naar rechts [Ac - ] e = (n Ac--n H3O+)/V en [HAc] e = (n HAc+n H3O+)/V • pH = -log K z.[HAc] e/[Ac - ] e (buffer)

Oefening buffer + sterk zuur / base • 50 ml buffer ([Ac - ] 0 = [HAc] 0 = 0,1 mol/l) + 0,1 ml [HCl] = 1 mol/l of + 1 ml [HCl] = 1 mol/l of + 0,1 ml [NaOH] = 1 mol/l of + 4 ml [NaOH] = 1 mol/l Bereken telkens de pH • Oefening 1, p. 147 • Vraagstukken p. 147: 1a , 2a (deel HCl en NaOH)

Oef<strong>en</strong>ing buffer + sterk zuur / base<br />

• 50 ml buffer ([Ac - ] 0 = [HAc] 0 = 0,1 mol/l)<br />

+ 0,1 ml [HCl] = 1 mol/l<br />

of + 1 ml [HCl] = 1 mol/l<br />

of + 0,1 ml [NaOH] = 1 mol/l<br />

of + 4 ml [NaOH] = 1 mol/l<br />

Berek<strong>en</strong> telk<strong>en</strong>s de pH<br />

• Oef<strong>en</strong>ing 1, p. 147<br />

• Vraagstukk<strong>en</strong> p. 147: 1a , 2a (deel HCl <strong>en</strong> NaOH)

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