CHEMISTRY: molar concentration question?

Emily

2016-05-23 3:57 pm

更新1:

100 mL of 0.12 M sodium hydroxide (NaOH) is added to 200 mL of 0.155 M hydrochloric acid (HCl). After neutralization, what is the concentration of HCl remaining in solution?

更新2:

Full answer step by step please!

回答 (1)

fooks

2016-05-23 4:08 pm

NaOH + HCl → NaCl + H₂O

Initial number of moles of NaOH = (0.12 mol/L) × (100/1000 L) = 0.012 mol
Initial number of moles of HCl = (0.155 mol/L) × (200/1000 L) = 0.031 mol > 0.012 mol

According to the equation, mole ratio NaOH : HCl = 1 : 1
Obviously, HCl is in excess, and NaOH is the limiting reactant (completely reacts).

Number of moles of HCl reacted = Number of moles of NaOH reacted = 0.012 mol
Number of moles of HCl remaining in the solution = (0.031 - 0.012) = 0.019 mol
Volume of the final solution = (100 + 200) mL = 300 mL = 0.3 L

Concentration of HCl remaining in the solution = (0.019 mol) / (0.3 L) = 0.0633 M

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