1. To find the mass of NaOH required to neutralize the HCl solution, we need to use the stoichiometry of the reaction between NaOH and HCl.
The balanced equation for the reaction is:
NaOH + HCl -> NaCl + H2O
From the equation, we can see that one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water.
First, let's calculate the number of moles of HCl in 300 mL of the 0.25 M solution:
Molarity (M) = moles/Liter
0.25 M = moles/0.3 L
moles = 0.25 M * 0.3 L
moles = 0.075 mol
Since the reaction is 1:1, we need 0.075 moles of NaOH to neutralize the HCl.
Now, we can calculate the mass of NaOH required using its molar mass:
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Mass of NaOH = moles * molar mass
Mass of NaOH = 0.075 mol * 40.00 g/mol
Mass of NaOH = 3.00 g
Therefore, you need to add 3.00 grams of NaOH to completely neutralize the 300 mL of HCl 0.25 M solution.
2. To find the concentration of H+ ions in the new solution, we can use the balanced equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
From the equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of water.
First, let's calculate the number of moles of HCl in the 100 mL solution containing 0.73 g of HCl:
Molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol
Moles of HCl = mass/molar mass
Moles of HCl = 0.73 g/36.46 g/mol
Moles of HCl = 0.02 mol
Since the reaction is 1:1, the number of moles of H+ ions produced is also 0.02 mol.
Now, let's calculate the concentration of H+ ions in the solution:
Concentration (M) = moles/Liter
Concentration (M) = 0.02 mol/0.1 L
Concentration (M) = 0.20 M
Therefore, the concentration of H+ ions in the new solution is 0.20 M.
3. To calculate the [H+] of the solution obtained by mixing 1 L of hydrochloric acid 1.0 M with 1 L of sodium hydroxide 0.990 M, we need to consider the balanced equation for the neutralization reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
From the balanced equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of water.
Since the solutions have equal volumes of 1 L, the moles of HCl and NaOH used will be equal.
The volume of the combined solution is 2 L, and the moles of H+ ions produced will also be equal to the moles of HCl and NaOH used.
Given that the concentration of hydrochloric acid is 1.0 M, the moles of HCl used is:
Moles of HCl = concentration * volume
Moles of HCl = 1.0 M * 1 L
Moles of HCl = 1.0 mol
Since the moles of HCl and NaOH used are equal, the moles of NaOH used is also 1.0 mol.
Now, let's calculate the concentration of H+ ions in the solution:
Concentration (M) = moles/volume
The total volume of the combined solution is 2 L, so the concentration of H+ ions is:
Concentration (M) = 1.0 mol/2 L
Concentration (M) = 0.50 M
Therefore, the [H+] of the solution obtained by mixing 1 L of hydrochloric acid 1.0 M with 1 L of sodium hydroxide 0.990 M is 0.50 M.