1.What mass of NaOH(s) must be added to 300 mL of HCl 0.25 M in order to completely neutralize this acid?

2.During a lab, you mix 2 solutions: a 100 ml solution containing 0.40 g of NaOH and a 100 mL solution containing 0.73 g of HCl. What is the concentration of H+ ions in the new solution?
3.Calculate the [H+] of a solution obtained by mixing 1 L of hydrochloric acid 1.0 M with 1 litre of sodium hydroxide 0.990 M.

1. mols HCl = M x L = ?

mols NaOH = mols HCl
mols NaOH = grams/molar mass. You know molar mass and mols, solve for grams.

2.mols NaOH = grams/molar mass = estimated 0.01
mols HCl = grams/molar mass = estimated 0.02.
NaOH + HCl ==> NaCl + H2O
0.02 mols HCl - 0.01 mols NaOH = 0.01 mols HCl unreacted.
M HCl = (H^+) = mols HCl/L solution. (L soln is 0.100 + 0.100 = ?)

3. mols HCl = M x L = 1.00
mols NaOH = M x L = 0.990
M HCl = (H^+) = (1.00-0.990)/2.00L = ?

Explain how you would prepare 0.25m of nitric acid in a 500ml volumetric flask

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.

1. To find the mass of NaOH needed to neutralize the HCl, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between NaOH and HCl.

The balanced chemical equation for the reaction is:
NaOH + HCl -> NaCl + H2O

From this equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl.

First, let's calculate the number of moles of HCl in 300 mL of 0.25 M HCl solution:

Number of moles of HCl = concentration (in moles/L) x volume (in L)
= 0.25 mol/L x 0.3 L
= 0.075 mol

Since the stoichiometry of the reaction is 1:1, we need 0.075 moles of NaOH to completely neutralize the acid.

To calculate the mass of NaOH needed, we need to know the molar mass of NaOH:
Na (22.99 g/mol) + O (16.00 g/mol) + H (1.01 g/mol) = 39.00 g/mol

Mass of NaOH = number of moles x molar mass
= 0.075 mol x 39.00 g/mol
= 2.93 g

Therefore, you would need to add approximately 2.93 grams of NaOH to completely neutralize the HCl.

2. To find the concentration of H+ ions in the new solution, we need to calculate the moles of HCl and the volume of the final solution.

The moles of HCl can be calculated using the formula:
moles = mass / molar mass

Molar mass of HCl = 1.01 g/mol for hydrogen + 35.45 g/mol for chlorine = 36.46 g/mol

moles of HCl = 0.73 g / 36.46 g/mol = 0.020 moles

Since the final solution has a total volume of 200 mL (100 mL + 100 mL), we can calculate the concentration of H+ ions:

Concentration of H+ ions = moles / volume (in L)
= 0.020 moles / 0.2 L
= 0.10 M

Therefore, the concentration of H+ ions in the new solution is 0.10 M.

3. To calculate the [H+] (concentration of H+ ions) of the solution obtained by mixing 1 L of hydrochloric acid (HCl) 1.0 M with 1 L of sodium hydroxide (NaOH) 0.990 M, we need to consider the stoichiometry of the reaction and the balanced chemical equation.

The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O

From this equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of water.

The initial moles of HCl in 1 L of a 1.0 M solution can be calculated using the formula:
moles = concentration (in moles/L) x volume (in L)

moles of HCl = 1.0 mol/L x 1 L = 1.0 mol

Since the stoichiometry of the reaction is 1:1, the moles of NaOH required to react completely with the HCl is also 1.0 mol.

The final volume of the solution obtained by mixing is 2 L (1 L + 1 L).

Therefore, the concentration of H+ ions in the final solution can be calculated as:
[H+] = moles / volume (in L)
= 1.0 mol / 2 L
= 0.50 M

Thus, the concentration of H+ ions in the solution obtained by mixing 1 L of HCl 1.0 M with 1 L of NaOH 0.990 M is 0.50 M.