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Volumul soluţiei de NaOH 4 M necesar neutralizării a 20 mL soluţie HCl 2 M este ?


Răspuns :

NaOH +  HCl => NaCl + HOH
V HCl = 20 mL = 0.02 L
C HCl = 2 M
C= n/V => n = V * C 
n HCl = 0.02 L * 2 mol/L = 0.04 mol
1 mol HCl ... 1 mol NaOH
0.04 mol HCL ... x mol NaOH 
x = 0.04 mol NaOH
C NaOH = 4 M
n NaOH = 0.04 mol
V= n/C => V NaOH = 0.04 mol / 4 mol/L => V NaOH = 0.01 L = 10 mL sol NaOH 4M