Answer:
a) 0.0379 M
b) 24.89 mL
c) 0.09 mL
Explanation:
The equation of the reaction is:
2HBr(aq) + Ca(OH)2(aq) ---> CaBr2(aq) + 2H20(l)
a)
Concentration of acid CA = 0.117 M
Concentration of base CB= ??
Volume of acid VA = 16.0 ml
Volume of base VB = 24.7 ml
number of moles of acid NA = 2
number of moles of base NB= 1
From;
CA VA/CB VB = NA/NB
CAVANB =CBVBNA
CB = CAVANB/VBNA
CB = 0.117 *16 * 1/ 24.7 *2
CB = 1.872/49.4
CB = 0.0379 M
b)
CA = 0.270 M
CB = 0.113 M
VA = ??
VB =22.8mL
NA =2
NB =1
From;
CA VA/CB VB = NA/NB
CAVANB =CBVBNA
VA = CBVBNA/CANB
VA = 0.113 * 22.8 * 2/0.207 *1
VA = 24.89 mL
c)
using the dilution formula
C1V1 =C2V2
Where:
C1= initial concentration =10.8 M
V1 = initial volume = ??
C2= final concentration =0.400 M
V2= final volume = 2.5 L
V1 = C2V2/C1
V1 = 0.400 * 2.5/10.8
V1 = 0.09 mL
