Answer to Question #319202 in General Chemistry for Emman

Question

To neutralize 15.8 mL of 0.687 molarity sodium hydroxide, how much 0.376 molarity HCl is needed? (5 points)

____HCl(aq)+____NaOH(aq)→____NaCl(aq)+____H(OH)(l)

Answer 1

HCl(aq) + NaOH(aq)→NaCl(aq) + H(OH)(l)

n(NaOH) = M(NaOH)×V(NaOH) = 0.687 mol/L × 0.0158 L = 0.01085 mol

n(NaOH) = n(HCl) = 0.01085 mol

V(HCl) = n(HCl) / M(HCl) = 0.01085 mol / 0.376 mol/L = 0.0288 l = 28.8 mL