Exercise 8 | Chemical Calculations for Solutions

In a titration experiment, it was found that 63.0 mL of 0.350 M KOH (aq) is required to neutralize 175 mL of HCl (aq).

What is the concentration of the HCl solution?

Equation: KOH (aq) + HCl (aq) → KCl (aq) + H2O (l)


According to the equation: 1 mole of KOH is required to neutralized 1 mole of HCl

⇒ when the endpoint of the neutralization is reached: nKOH = nHCl

⇒ MKOH x VKOH = MHCl x VHCl

⇒ MHCl = MKOH × VKOHVHCl

⇒ MHCl = 0.350 × 63.0175 = 0.126 M