An acid solution is 0.100 mol L^-1 in HCl and 0.200 mol L^-1 . What volume of a 0.150 mol L^-1 KOH solution would completely neutralize 500.0 mL of this solution? A) 0.700 L B) 1.98 L C) 1.67 L D) 1.02 L E) 0.250 L

Question

Quizplus · Accepted Answer

The balanced chemical equation for the neutralization reaction between HCl and KOH is: HCl + KOH → H2O + KCl From the equation, we see that one mole of HCl reacts with one mole of KOH to produce one mole of water and one mole of KCl. Therefore, the number of moles of HCl in the solution is the same as the number of moles of KOH required to neutralize it. The number of moles of HCl in 500.0 mL of the solution can be calculated as: 0.100 mol L^-1/L × 0.5000 L = 0.050 mol L^-1 The number of moles of KOH required to neutralize the solution is therefore also 0.050 mol L^-1. To calculate the volume of the 0.150 mol L^-1 KOH solution required, we can use the formula: moles of solute = concentration of solution × volume of solution Solving for volume, we get: volume of solution = moles of solute / concentration of solution Plugging in the values, we get: volume of KOH solution = 0.050 mol L^-1 / 0.150 mol L^-1/L = 0.333 L Converting the volume to liters, we get: volume of KOH solution = 0.333 L = 1.67 L Therefore, the answer is choice C.

An Acid Solution Is 0