The question is incomplete, as it does not specify what the hydrochloric acid reacts with. Based on the context of previous questions involving KOH and the number "60" visible, we will assume

The question is incomplete, as it does not specify what the hydrochloric acid reacts with. Based on the context of previous questions involving $\text{KOH}$ and the number "60" visible, we will assume the question asks for the volume of $\text{HCl}$ required to react with $60 \text{ cm}^3$ of the $\text{KOH}$ solution from Question 6(a)(ii).

• Assumed Given:
  • Concentration of acid ($\text{HCl}$), $C_A = 0.15 \text{ moldm}^{-3}$
  • Volume of base ($\text{KOH}$), $V_B = 60 \text{ cm}^3$
  • Concentration of base ($\text{KOH}$), $C_B = 0.0798 \text{ moldm}^{-3}$ (from Question 6(a)(ii))
  • Balanced equation: $\text{HCl}_{(aq)} + \text{KOH}_{(aq)} \longrightarrow \text{KCl}_{(aq)} + \text{H}_2\text{O}_{(l)}$
  • Mole ratio of $\text{HCl} : \text{KOH}$ is $n_A : n_B = 1:1$.