Consider the reaction represented by the question below: KOH(aq) + HCI(aq) → KCI(aq) + H2O(l) What volume of 0.25 mol dm-3 KOH would be required to complete...

Consider the reaction represented by the question below: KOH_(aq) + HCI_(aq) → KCI_(aq) + H₂O_(l) What volume of 0.25 mol dm^-3 KOH would be required to completely neutralize 40 cm³ of 0.10 mol dm ^-3 HCI?

Answer Details

The balanced equation for the reaction between KOH and HCl is: KOH + HCl → KCl + H₂O From the equation, we can see that the mole ratio of KOH to HCl is 1:1. This means that for every 1 mole of HCl, 1 mole of KOH is required to neutralize it. To find the volume of 0.25 mol dm^-3 KOH needed to neutralize 40 cm³ of 0.10 mol dm^-3 HCl, we need to use the formula: n_acid V_acid = n_base V_base where n is the number of moles and V is the volume in dm³. Rearranging the formula to isolate V_base, we get: V_base = (n_acid V_acid) / n_base Substituting the values we have: n_acid = 0.10 mol dm^-3 (concentration of HCl) V_acid = 40 cm³ = 0.04 dm³ (volume of HCl) n_base = 0.25 mol dm^-3 (concentration of KOH) V_base = (0.10 mol dm^-3 x 0.04 dm³) / 0.25 mol dm^-3 = 0.016 dm³ = 16 cm³ Therefore, the volume of 0.25 mol dm^-3 KOH needed to completely neutralize 40 cm³ of 0.10 mol dm^-3 HCl is 16 cm³. The answer is (d) 16cm³.