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23.74cm3 0.0147 mol dm-3NaOHare required to neutralize 25.00 cm3 of HCl calculate concentration of HCl acid in moldm-3and g dm-3?
Updated: 4/28/2022
Wiki User
∙ 12y agoBest Answer
HCl + NaOH ― NaCl + H2O
nNaOH = c*v
= 0.01470 * 23.74*10-3L
= 3.49*10-4 mol
1 mol HCl reacts with 1 mol NaOH
therefore
nHCl = 3.49*10-4mol
CHCL= nNaOH /VNaOH
= 3.49*10-4 / 25.00*10-3
= 13.96 * 10-3 mol.dm-3
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∙ 12y agoThis answer is:
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