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?

Question

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?

| General Chemistry 2

Accepted Answer

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: n_KOH = n_HCl

⇒ M_KOH x V_KOH = M_HCl x V_HCl

⇒ M_HCl = $\frac{M_{K O H} \times V_{KOH}}{V_{HCl}}$

⇒ M_HCl =

\frac{0.350 \times 63.0}{175}

= 0.126 M

Acid-base titrations - 2 | Chemical Calculations