Kimia Sekolah Menengah Pertama 400 cm of sodium hydroxide solution 0.5 mol dm 3 is used to neutralise 100 cm of hydrochloric
acid solution as shown in the reaction below.
NaOH(aq) + HCl(aq) - NaCl(aq) + H2O(1)
4
What is the concentration of the hydrochloric acid solution?
A 0.100 mol dm3
С 1.00 mol dm3
B 0.200 mol dm3
D 2.00 mol dm3

By kimberlybednarulv di
400 cm of sodium hydroxide solution 0.5 mol dm 3 is used to neutralise 100 cm of hydrochloric acid solution as shown in the reaction below. NaOH(aq) + HCl(aq) - NaCl(aq) + H2O(1) 4 What is the concentration of the hydrochloric acid solution? A 0.100 mol dm3 С 1.00 mol dm3 B 0.200 mol dm3 D 2.00 mol dm3 Jawaban: D. 2,00 mol/dm³ Penjelasan: V NaOH = 400 cm³ = 0,4 dm³ M NaOH = 0,5 mol/dm³ or 0,5 M V HCl = 100 cm³ = 0,1 dm³ Reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Ask: concentration of HCl? answer: neutralise equation is: Va x Ma x a = Vb x Mb x b 0,1 dm³ x Ma x 1 = 0,4 dm³ x 0,5 M x 1 0,1 dm³ x Ma = 0,2 mol Ma = 0,2 mol / 0,1 dm³ = 2,00 M or 2,00 mol/dm³ nb: cm³ = ml dm³ = L cmiiw ^^