how many drops of 0.5M NaOH are needed to neutralize 10 drops of 1.0M HCl? What would be the pH of the neutral solution?
drops x M = drops x M
HCl + NaOH ==> NaCl + H2O
The pH of the solution is determined by the salt; in this case, neither the Na^+ nor the Cl^- are hydrolyzed; therefore, the pH will be the pH of pure water.
H2O ==> H^+ + OH^-
(H^+)(OH^-) = Kw = 1 x 10^-14
Solve for (H^+)
1 drop
To determine how many drops of 0.5M NaOH are needed to neutralize 10 drops of 1.0M HCl, we can use the concept of stoichiometry and the balanced chemical equation for the reaction.
The balanced equation for the reaction between NaOH and HCl is:
NaOH + HCl ➞ NaCl + H2O
From the equation, we can see that the stoichiometric ratio between NaOH and HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.
First, let's calculate the moles of HCl involved in 10 drops of 1.0M HCl.
Molarity (M) is defined as moles of solute per liter of solution. Therefore, we need to find the moles of HCl in 10 drops.
The volume of one drop is not provided in the question, so we assume it to be negligible for simplicity.
Moles of HCl = Molarity × Volume (in liters)
= 1.0 mol/L × 10 drops × (1 L / 1000 drops)
= 0.010 mol
Since the stoichiometric ratio between NaOH and HCl is 1:1, we need an equal number of moles of NaOH to neutralize the HCl. Therefore, we need 0.010 moles of NaOH.
Now, let's determine the volume of 0.5M NaOH required to contain 0.010 moles.
Volume (in liters) = Moles / Molarity
= 0.010 mol / 0.5 mol/L
= 0.020 L
Since 1 liter contains 1000 milliliters (mL), the volume comes out to be:
Volume (in mL) = 0.020 L × (1000 mL / 1 L)
= 20 mL
Therefore, 20 drops of 0.5M NaOH are needed to neutralize 10 drops of 1.0M HCl.
To determine the pH of the neutral solution after the reaction, we need to consider the concentration of the resulting salt (NaCl). Since NaCl is a strong electrolyte, it completely dissociates into Na+ and Cl- ions in solution.
The pH of a neutral solution is 7.