Question

Cantităţi egale de acid palmitic, respectiv stearic necesită pentru neutralizare:
A) cantităţi egale de KOH
B) acelaşi volum de soluţie NaOH10%
C) aceeaşi cantitate de NaCl
D) acelaşi număr de moli NaOH
E) acelaşi volum de soluţie KOH10%
F) număr diferit de moli NaOH

Cu explicatie!

Answer 1

neutralizare = acid + baza
acid stearic: CH3 - (CH2)16 - COOH
acid palmitic: CH3 - (CH2)14 - COOH
CH3 - (CH2)16 - COOH + KOH = CH3 - (CH2)16 - COOK + H2O
CH3 - (CH2)14 - COOH + KOH = CH3 - (CH2)14 - COOK + H2O
CH3 - (CH2)16 - COOH + NaOH = CH3 - (CH2)16 - COONa + H2O
CH3 - (CH2)14 - COOH + NaOH = CH3 - (CH2)14 - COONa + H2O
Fie 1 mol din fiecare acid (poti lua cat vrei, nu se

1 mol acid stearic (din reactie).....1 mol NaOH
1 mol acid stearic (presupus)..........x moli NaOH
x = 1 mol = 40g NaOH
c%=md/Vs * 100=> Vs = 40/10*100= 400 ml

1 mol acid palmitic (din reactie).....1 mol NaOH
1 mol acid palmitic (presupus)..........x moli NaOH
x = 1 mol = 40g NaOH
c%=md/Vs * 100=> Vs = 40/10*100= 400 ml

1 mol acid stearic (din reactie).....1 mol KOH
1 mol acid stearic (presupus)..........x moli KOH
x = 1 mol = 56 g NaOH
c%=md/Vs * 100=> Vs = 56/10*100= 560 ml

1 mol acid palmitic (din reactie).....1 mol KOH
1 mol acid palmitic (presupus)..........x moli KOH
x = 1 mol = 56g KOH
c%=md/Vs * 100=> Vs = 56/10*100= 560ml

Raspunsuri corecte: a, b, d, e