1) How many drops of 12M HCl would you add to 1.00 L of 0.100 M HC2H3O2 to make [C2H3O2-] = 1.0 x 10^-4 M? Assume that 1 drop = 0.050 mL and that the volume of the solution remains 1.00 L after the 12 M HCl is diluted. [Hint: What must be the [H3O+] in the solution?]

2. What mass of NaC2H3O2 should be added to 1.00 L of 0.100 M HC2H3O2 to produce a solution with pH = 5.00? Assume that the volume reamins 1.00 L.

I accidentally deleted my response. Let me know if you didn't get it and I will repost.

Can you repost #1 please. Thanks.

Let't call acetic acid, HC2H3O2, HAc.

HAc ==> H^+ + Ac^-
Ka = (H^+)(Ac^-)/(HAc)
Ka is about 1.8 x 10^-5 but you need to verify that from your text.
You know Ka, (Ac^-) is 1 x 10^-4 from the problem and (HAc) is 0.1 from the problem. Calculate (H+) that is needed.
Since you have a liter of solution, the moles H^+ (moles/L) will be the same as the molarity.
Now determine how much of the 12 M HCl must be added to make the H^+ from above.
12 M x volume = moles H^+.
Then knowing that each drop is 0.05 mL, convert volume to drops.
I played around with some numbers in my head and came up with approximately 6 mL and 120 drops BUT check my work. I just estimated here and there. You MUST do it yourself to confirm.
The second problem is one to solve using the Henderson-Hasselbalch equation.

1) Well, to find out how many drops of 12M HCl you would need, you first have to determine what my HClometer will say. Just kidding! Let's calculate it step by step.

The balanced equation for the reaction is:

HC2H3O2 (aq) + HCl (aq) -> H3O+ (aq) + C2H3O2- (aq)

From the equation, we can see that one mole of HC2H3O2 reacts with one mole of HCl to produce one mole of C2H3O2-. Therefore, to make [C2H3O2-] = 1.0 x 10^-4 M, an equal amount of moles of HCl is needed.

Given that the initial concentration of HC2H3O2 is 0.100 M and the final concentration of C2H3O2- is 1.0 x 10^-4 M, the change in concentration is 0.100 M - 1.0 x 10^-4 M = 0.0999 M.

Now we need to find out how many moles of HCl are present in this change in concentration.

1 mole of HC2H3O2 reacts with 1 mole of HCl, so the number of moles of HCl needed is also 0.0999 M.

Given that the volume of the solution is 1.00 L, we can use the equation C = n/V to find out the number of moles of HCl needed.

0.0999 M = n/1.00 L

n = 0.0999 moles

Now, we need to find out the number of drops of 12M HCl needed. Given that 1 drop = 0.050 mL, we can calculate:

Number of drops = (0.0999 moles)(1000 mL/12 moles)(1 drop/0.050 mL)

Number of drops = 1398.33 drops

So, you would need approximately 1398 drops of 12M HCl to make [C2H3O2-] = 1.0 x 10^-4 M. Just don't drop the ball!

2) How much NaC2H3O2 do you need to make a solution with pH = 5.00? Well, this one is a bit tricky, but let's shake it up!

First, we need to determine the [H3O+] concentration in the solution with a pH of 5.00. Since pH is defined as the negative logarithm of the [H3O+] concentration, we can calculate it:

pH = -log[H3O+]

5.00 = -log[H3O+]

[H3O+] = 10^-5 M

From the balanced equation:

HC2H3O2 (aq) + H2O (l) -> H3O+ (aq) + C2H3O2- (aq)

We can see that one mole of HC2H3O2 reacts with one mole of H3O+ to produce one mole of C2H3O2-. Therefore, to produce a solution with [H3O+] = 10^-5 M, we need an equal number of moles of HC2H3O2 and H3O+.

Given that the volume of the solution is 1.00 L, we can use the equation C = n/V to find out the number of moles of HC2H3O2 needed.

10^-5 M = n/1.00 L

n = 10^-5 moles

Now, we need to find out the mass of NaC2H3O2 needed. Given the molar mass of NaC2H3O2 is approximately 82.03 g/mol, we can calculate:

Mass of NaC2H3O2 = (10^-5 moles)(82.03 g/mol)

Mass of NaC2H3O2 = 0.0008203 g

So, you would need approximately 0.0008203 g of NaC2H3O2 to add to 1.00 L of 0.100 M HC2H3O2 to produce a solution with pH = 5.00. That's a pinch of salt-joke!

To solve these problems, we need to use some concepts from chemistry, such as the dilution formula, the equation for pH, and stoichiometry. Let's break down each question and explain how to find the answer.

1) How many drops of 12M HCl would you add to 1.00 L of 0.100 M HC2H3O2 to make [C2H3O2-] = 1.0 x 10^-4 M? Assume that 1 drop = 0.050 mL and that the volume of the solution remains 1.00 L after the 12 M HCl is diluted.

To find the number of drops required to achieve the desired concentration, we need to first determine the new concentration of the acetic acid (HC2H3O2) after it reacts with the hydrochloric acid (HCl).

1. Use the stoichiometry of the reaction:

HC2H3O2 + HCl → H3O+ + C2H3O2-

From the reaction, we can see that the mole ratio between HC2H3O2 and C2H3O2- ions is 1:1. Therefore, when HC2H3O2 reacts completely, it will produce the same amount of C2H3O2- ions.

2. Calculate the number of moles of C2H3O2- needed to achieve the desired concentration:

[C2H3O2-] = 1.0 x 10^-4 M

Volume of the solution = 1.00 L

Number of moles of C2H3O2- = [C2H3O2-] × Volume
= (1.0 x 10^-4 M) × (1.00 L)
= 1.0 x 10^-4 moles

3. Determine the volume of 12 M HCl required to react with the moles of C2H3O2- ions:

Using the stoichiometry of the reaction and the mole ratio, we know that each mole of HC2H3O2 reacts with 1 mole of HCl to produce 1 mole of C2H3O2-.

Volume of 12 M HCl = (Number of moles of C2H3O2-) ÷ (Concentration of HCl)
= (1.0 x 10^-4 moles) ÷ (12 M)
= 8.33 x 10^-6 L

4. Convert the volume of HCl to the number of drops:

Conversion factor: 1 drop = 0.050 mL

Number of drops = (Volume of HCl in L) ÷ (Conversion factor)
= (8.33 x 10^-6 L) ÷ (0.050 mL)
= 1.67 x 10^-4 drops

Therefore, you would need to add approximately 1.67 x 10^-4 drops (or round it to the nearest whole number) of 12 M HCl to the solution.

2) What mass of NaC2H3O2 should be added to 1.00 L of 0.100 M HC2H3O2 to produce a solution with pH = 5.00? Assume that the volume remains 1.00 L.

To solve this problem, we need to determine the amount of sodium acetate (NaC2H3O2) needed to adjust the pH of the acetic acid (HC2H3O2) solution.

1. Calculate the concentration of H3O+ ions from the pH value:

pH = -log[H3O+]
5.00 = -log[H3O+]
[H3O+] = 10^-5.00 M
= 1.0 x 10^-5 M

2. Use the equation for acetic acid dissociation to determine the concentration of C2H3O2- ions:

[H+] = [C2H3O2-]
1.0 x 10^-5 M = [C2H3O2-]

3. Determine the number of moles of C2H3O2- needed to achieve this concentration:

Number of moles of C2H3O2- = [C2H3O2-] × Volume
= (1.0 x 10^-5 M) × (1.00 L)
= 1.0 x 10^-5 moles

4. Convert the moles of C2H3O2- to grams of NaC2H3O2 using the molar mass:

Molar mass of NaC2H3O2 = (22.99 g/mol) + (12.01 g/mol) + (3 x 1.01 g/mol) + (2 x 16.00 g/mol)
= 82.03 g/mol

Mass of NaC2H3O2 = (Number of moles of C2H3O2-) × (Molar mass of NaC2H3O2)
= (1.0 x 10^-5 moles) × (82.03 g/mol)
= 8.20 x 10^-4 g

Therefore, you would need to add approximately 8.20 x 10^-4 grams (or round it to the appropriate precision) of NaC2H3O2 to the solution.