1) A 25mL sample of the .265M HCI solution from the previous question is

titrated with a solution of NaOH. 28.25mL of the NaOH solution is required
to titrate the HCl. Calculate the molarity of the NaOH solution.

2) A 1.12g sample of an unknown
monoprotic acid is titrated with 29.15mL of .315M NaOH. Calculate the molar mass of the acid.

3) A student dissolves a 1.18g aspirin tablet in water and titrates the
solution with a .265M NaOH. 22.05mL of NaOH are required to reach the
equivalence point. Calculate the percentage of acetylsalicylic acid in the tablet.

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The titration equation

Heqacid*Molarityacid*volumeacid=Heqbase*Molaritybase*volumebase

where Heqacid=1 for monoprotic acid, 2 for diprotic, and 3 for triprotic acid
Heqbase=1 for single OH, 2 for two OH, and 3 for three OH

for your first question: first get a balanced equation, then calculate moles of HCl from that, then use the equation to get moles of NaOH and you use given information to get M NaOH

for your third question try using the equation (mass of acid/mass of tablet)x 100

1) To solve this problem, we can use the concept of stoichiometry. The balanced equation for the reaction between HCl and NaOH is:

HCl + NaOH -> NaCl + H2O

From the equation, we can see that the ratio between HCl and NaOH is 1:1. Therefore, the number of moles of HCl present in the solution can be calculated using the formula:

moles of HCl = Molarity of HCl * Volume of HCl solution

moles of HCl = 0.265 M * 0.025 L
moles of HCl = 0.006625 mol

Since the stoichiometric ratio between HCl and NaOH is 1:1, the number of moles of NaOH required to react with HCl is also 0.006625 mol.

Now, let's calculate the molarity of NaOH solution:

Molarity of NaOH = Moles of NaOH / Volume of NaOH solution

Molarity of NaOH = 0.006625 mol / 0.02825 L
Molarity of NaOH = 0.234 M

Therefore, the molarity of the NaOH solution is 0.234 M.

2) To determine the molar mass of the unknown acid, we need to first find the number of moles of NaOH used in the titration. The balanced equation for the reaction between the acid and NaOH is:

Acid + NaOH -> Salt + Water

From the equation, we can see that the stoichiometric ratio between the acid and NaOH is 1:1. Therefore, the number of moles of NaOH used can be calculated using the formula:

moles of NaOH = Molarity of NaOH * Volume of NaOH solution

moles of NaOH = 0.315 M * 0.02915 L
moles of NaOH = 0.00917225 mol

Since the stoichiometric ratio between the acid and NaOH is 1:1, the number of moles of the unknown acid is also 0.00917225 mol.

Now, let's calculate the molar mass of the unknown acid:

Molar mass of acid = Mass of acid / Moles of acid

Molar mass of acid = 1.12 g / 0.00917225 mol
Molar mass of acid ≈ 122.06 g/mol

Therefore, the molar mass of the unknown acid is approximately 122.06 g/mol.

3) To determine the percentage of acetylsalicylic acid in the tablet, we need to first find the number of moles of NaOH used in the titration. The balanced equation for the reaction between acetylsalicylic acid and NaOH is:

Acetylsalicylic acid + NaOH -> Salt + Water

From the equation, we can see that the stoichiometric ratio between the acetylsalicylic acid and NaOH is 1:1. Therefore, the number of moles of NaOH used can be calculated using the formula:

moles of NaOH = Molarity of NaOH * Volume of NaOH solution

moles of NaOH = 0.265 M * 0.02205 L
moles of NaOH = 0.00583725 mol

Since the stoichiometric ratio between acetylsalicylic acid and NaOH is 1:1, the number of moles of acetylsalicylic acid in the tablet is also 0.00583725 mol.

Now, let's calculate the percentage of acetylsalicylic acid in the tablet:

Percentage of acetylsalicylic acid = (Molar mass of acetylsalicylic acid / Molar mass of tablet) * 100%

The molar mass of acetylsalicylic acid (C9H8O4) can be calculated as:

C: 9 * 12.01 g/mol = 108.09 g/mol
H: 8 * 1.01 g/mol = 8.08 g/mol
O: 4 * 16.00 g/mol = 64.00 g/mol

Molar mass of acetylsalicylic acid = 108.09 g/mol + 8.08 g/mol + 64.00 g/mol
Molar mass of acetylsalicylic acid ≈ 180.17 g/mol

Now, let's calculate the percentage of acetylsalicylic acid in the tablet:

Percentage of acetylsalicylic acid = (0.00583725 mol / 0.00118 g) * 100%
Percentage of acetylsalicylic acid ≈ 494.94%

Therefore, the percentage of acetylsalicylic acid in the tablet is approximately 494.94%.