Home many drops of 0.1M HCl is needed to neutralize 10mL of 1M NaOH?
How many drops are in 1mL? Most of my pipets are about 15 drops/ml, but you have to know drop size in your question.
MolarityHCl*volumeinMl=10*1
volume in drops=10/.1 * XXdrops/ml
32
To determine how many drops of 0.1M HCl are needed to neutralize 10mL of 1M NaOH, we can use the concept of stoichiometry and the equation of neutralization. The balanced equation for the reaction between HCl and NaOH is:
HCl + NaOH ⟶ NaCl + H2O.
From the equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of water.
Given that the concentration of HCl is 0.1M, we can calculate how many moles of HCl are present in 10mL (0.01L) of the solution:
moles of HCl = volume (L) × concentration (M)
= 0.01L × 0.1M
= 0.001 moles.
Since the stoichiometry of the reaction is 1:1 between HCl and NaOH, we know that we need 0.001 moles of HCl to react with 0.001 moles of NaOH.
To convert moles to drops, we need to know the drop size, which can vary. Assuming a drop size of approximately 0.05mL, we can calculate the number of drops as follows:
drops = moles × (1L / 0.05mL)
= 0.001 × (1L / 0.05mL)
= 20 drops.
Therefore, approximately 20 drops of 0.1M HCl are needed to neutralize 10mL of 1M NaOH, assuming a drop size of 0.05mL.