how many ml of a 2M solution of HCl would be needed to neutrilize 2ml of a 5M solution of KOH?
To determine the number of milliliters (ml) of a 2M solution of HCl needed to neutralize 2ml of a 5M solution of KOH, we need to use the concept of stoichiometry and the balanced chemical equation of the neutralization reaction between HCl and KOH:
HCl + KOH → H2O + KCl
The balanced equation indicates that one mole of HCl reacts with one mole of KOH to produce one mole of water and one mole of KCl.
First, we need to calculate the number of moles of KOH present in 2ml of a 5M solution. Since the molarity (M) is defined as moles of solute per liter of solution, we can use the formula:
moles = concentration (M) × volume (L)
Given that the volume is 2ml (0.002L) and the concentration is 5M for KOH, we can calculate the number of moles of KOH:
moles of KOH = 5M × 0.002L = 0.01 moles
Since the stoichiometry of the reaction tells us that the reaction is 1:1, we know that 0.01 moles of KOH will react with 0.01 moles of HCl.
Now, we can calculate the volume of the 2M HCl needed to neutralize 0.01 moles of HCl using the formula:
volume (L) = moles / concentration (M)
Given that the concentration is 2M for HCl, we can calculate the volume of HCl:
volume of HCl = 0.01 moles / 2M = 0.005L
Finally, we convert the volume from liters to milliliters:
volume of HCl = 0.005L × 1000ml/L = 5ml
Therefore, 5ml of the 2M HCl solution would be needed to neutralize 2ml of the 5M KOH solution.