What volume of a .500 M HCl solution is needed to neutralize each of the following:

10 ml of a .300 M NaOH solution
10 mL of a .200 M Ba(OH)2 solution

Why did the acid go to the party? Because it was looking for some base to spend some quality time with!

To find the volume of HCl solution needed to neutralize each of the given solutions, we can use the concept of stoichiometry from the balanced chemical equations of the neutralization reactions.

For the first case, reacting HCl with NaOH, the balanced equation is:

HCl + NaOH → NaCl + H2O

According to the balanced equation, the mole ratio between HCl and NaOH is 1:1. This means that for every mole of NaOH, we will need 1 mole of HCl to neutralize it.

Now, let's calculate the moles of NaOH in the 10 mL of a 0.300 M NaOH solution:

Molarity (M) = Moles/Liters

0.300 M = Moles/0.010 L

Moles = 0.300 M x 0.010 L = 0.003 moles

Since the mole ratio is 1:1, we would need 0.003 moles of HCl to neutralize the NaOH.

Next, we need to find the volume of a 0.500 M HCl solution containing 0.003 moles of HCl:

Moles = Molarity x Volume

0.003 moles = 0.500 M x Volume

Volume = 0.003 moles / 0.500 M

Volume = 0.006 L = 6 mL

Therefore, approximately 6 mL of a 0.500 M HCl solution is needed to neutralize 10 mL of a 0.300 M NaOH solution.

Let's move on to the second case with Ba(OH)2. Shall we?

Why did the base bring a towel to the party? In case things got a little too basic!

The balanced equation for the reaction of HCl with Ba(OH)2 is:

2HCl + Ba(OH)2 → 2H2O + BaCl2

This tells us that the mole ratio between HCl and Ba(OH)2 is 2:1. So, for every 2 moles of HCl, we will need 1 mole of Ba(OH)2 to neutralize it.

We can start by calculating the moles of Ba(OH)2 in the 10 mL of a 0.200 M Ba(OH)2 solution:

Moles = Molarity x Volume

0.200 M x 0.010 L = 0.002 moles

Since the mole ratio is 2:1, we would need half the number of moles of HCl, which is 0.001 moles.

Now, let's find the volume of a 0.500 M HCl solution containing 0.001 moles of HCl:

Volume = Moles / Molarity

Volume = 0.001 moles / 0.500 M

Volume = 0.002 L = 2 mL

So, approximately 2 mL of a 0.500 M HCl solution is needed to neutralize 10 mL of a 0.200 M Ba(OH)2 solution.

By the way, be careful while handling those strong acids and bases. I'm here to make chemistry a little less scary, not add to the chaos!

To determine the volume of HCl solution needed to neutralize each of the given solutions, we can use the concept of stoichiometry and the equation:

HCl + NaOH → NaCl + H2O (for the first problem)
HCl + Ba(OH)2 → BaCl2 + 2H2O (for the second problem)

Let's calculate the volume of HCl solution needed step by step:

First problem: Neutralizing NaOH solution (0.300 M)
From the balanced equation, we can see that 1 mole of HCl will react with 1 mole of NaOH. This means the ratio of moles of HCl to NaOH is 1:1.

1. Calculate the number of moles of NaOH:
Moles of NaOH = concentration x volume
Moles of NaOH = 0.300 M x 0.010 L (convert 10 mL to liters)
Moles of NaOH = 0.003 moles

2. Since the ratio of HCl to NaOH is 1:1, we need 0.003 moles of HCl to neutralize the NaOH solution.

3. Calculate the volume of HCl solution needed:
Moles of HCl = 0.003 moles
Concentration of HCl = 0.500 M

Volume of HCl = Moles of HCl / Concentration of HCl
Volume of HCl = 0.003 moles / 0.500 M
Volume of HCl = 0.006 L (convert to mL)
Volume of HCl = 6 mL

Therefore, 6 mL of a 0.500 M HCl solution is needed to neutralize 10 mL of a 0.300 M NaOH solution.

Second problem: Neutralizing Ba(OH)2 solution (0.200 M)
From the balanced equation, we can see that 1 mole of HCl will react with 1 mole of Ba(OH)2. This means the ratio of moles of HCl to Ba(OH)2 is 1:1.

1. Calculate the number of moles of Ba(OH)2:
Moles of Ba(OH)2 = concentration x volume
Moles of Ba(OH)2 = 0.200 M x 0.010 L (convert 10 mL to liters)
Moles of Ba(OH)2 = 0.002 moles

2. Since the ratio of HCl to Ba(OH)2 is 1:1, we need 0.002 moles of HCl to neutralize the Ba(OH)2 solution.

3. Calculate the volume of HCl solution needed:
Moles of HCl = 0.002 moles
Concentration of HCl = 0.500 M

Volume of HCl = Moles of HCl / Concentration of HCl
Volume of HCl = 0.002 moles / 0.500 M
Volume of HCl = 0.004 L (convert to mL)
Volume of HCl = 4 mL

Therefore, 4 mL of a 0.500 M HCl solution is needed to neutralize 10 mL of a 0.200 M Ba(OH)2 solution.

To find the volume of the HCl solution required to neutralize a given volume of the base solution, we can use the concept of stoichiometry.

Let's start by writing the balanced chemical equations for the neutralization reactions:
1. HCl + NaOH -> NaCl + H2O
2. 2HCl + Ba(OH)2 -> BaCl2 + 2H2O

From the equations, we can see that one mole of HCl reacts with one mole of NaOH and two moles of HCl react with one mole of Ba(OH)2.

First, let's determine the amount of moles of base in each case:
1. For the first case, 10 mL of a 0.300 M NaOH solution contains:
Moles of NaOH = Volume (L) x Concentration (M)
= 0.010 L x 0.300 M
= 0.003 moles

2. For the second case, 10 mL of a 0.200 M Ba(OH)2 solution contains:
Moles of Ba(OH)2 = Volume (L) x Concentration (M)
= 0.010 L x 0.200 M
= 0.002 moles

Now, let's determine the volume of the HCl solution needed for each case.

1. To neutralize 0.003 moles of NaOH, we require an equal number of moles of HCl. Since one mole of NaOH reacts with one mole of HCl, we need 0.003 moles of HCl.

Concentration (M) = Moles / Volume (L)
Volume (L) = Moles / Concentration (M)
Volume (L) = 0.003 moles / 0.500 M
Volume (L) = 0.006 L

So, 0.006 L or 6 mL of a 0.500 M HCl solution is needed to neutralize 10 mL of a 0.300 M NaOH solution.

2. To neutralize 0.002 moles of Ba(OH)2, we require twice the number of moles of HCl. Since two moles of HCl react with one mole of Ba(OH)2, we need 2 * 0.002 = 0.004 moles of HCl.

Volume (L) = Moles / Concentration (M)
Volume (L) = 0.004 moles / 0.500 M
Volume (L) = 0.008 L

So, 0.008 L or 8 mL of a 0.500 M HCl solution is needed to neutralize 10 mL of a 0.200 M Ba(OH)2 solution.

Please remember to include the leading zero when writing values.

Start with the equation for the neutralisation

HCl + NaOH -> NaCl + H2O

so one mole of HCl neutralises 1 mole of NaOH

we have

10 x 0.300 mmoles of NaOH so we need the same number of mmoles of HCl

so if V is the volume of HCl

10 x 0.300 = V x 0.500

solve for V for the volume of HCl

repeat for Ba(OH)2