0 05 mol dm-3 HCl is neutralized by 25 cm3 NaOH... - JAMB Chemistry 2008 Question
0.05 mol dm-3 HCl is neutralized by 25 cm3 NaOH. If the volume of acid used is 32.00cm3, what is the concentration of the base
A
0.016 mol dm-3
B
0.032 mol dm-3
C
0.039 mol dm-3
D
0.064 mol dm-3
correct option: d
HCl + NaOH → NaCl + H2O
CAVA = NA/NB = (0.05 * 32)/(CB * 25)
∴ CB = (0.05 * 32)/25 = 0.064 mol dm-3
CAVA = NA/NB = (0.05 * 32)/(CB * 25)
∴ CB = (0.05 * 32)/25 = 0.064 mol dm-3
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