Question

For the following acids of varying concentrations, which are titrated with 0.125 M KOH, rank the acids in order of least to most volume of base needed to completely neutralize the acid. 0.060 M H2SO3 (diprotic) 0.075 M HNO2 (monoprotic) 0.060 M H3AsO4 (triprotic) 0.15 M HC2H3O2 (monoprotic)

0.095 M H2C6H6O6 (diprotic)
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least volume of base most volume of base

Answer 1

From most to least: 0.095M H2C6H6O6 > 0.060M H3AsO4 > 0.15M HC2H3O2 > 0.060M H2SO3 > 0.075M HNO2

Step-by-step explanation:

The easiest way to tackle this problem is to use the equation ViMi = MfVf, which states that the initial volume times the initial molarity equals the final volume times the final molarity. Since they don't give us an initial volume and cause it's constant, you can assign one; I chose 25mL. Also, based on if it's monoprotic -> triprotic, you need to divide that 0.125 by that number.

(0.095M H2C6H6O6)(25mL) = (0.125 M)(1/2x) = 38mL

(0.060M H3AsO4)(25mL) = (0.125 M)(1/3x) = 36mL

(0.15M HC2H3O2)(25mL) = (0.125 M)(x) = 30 mL

(0.060M H2SO3)(25mL) = (0.123 M)(1/2x) = 24mL

(0.075M HNO2)(25mL) = (0.123 M)(x) = 15mL

Answer 2

The ranking from least to most volume of base needed is
`\(\ce{HNO2} < \ce{HC2H3O2} < \ce{H2SO3} < \ce{H2C6H6O6} < \ce{H3AsO4}\)`.

To rank the acids in order of least to most volume of base needed to completely neutralize the acid when titrated with 0.125 M KOH, we can consider the molarity and the number of acidic hydrogens (protons) each acid can donate.

Here are the acids along with their molarities and the number of acidic hydrogens:

1.
`\(0.060 \, \text{M} \, \ce{H2SO3}\)` (diprotic)

2.
`\(0.075 \, \text{M} \, \ce{HNO2}\)` (monoprotic)

3.
`\(0.060 \, \text{M} \, \ce{H3AsO4}\)` (triprotic)

4.
`\(0.15 \, \text{M} \, \ce{HC2H3O2}\)` (monoprotic)

5.
`\(0.095 \, \text{M} \, \ce{H2C6H6O6}\)` (diprotic)

The volume of base needed is directly proportional to the number of moles of acid, which is the product of the molarity and the volume of the solution.

Additionally, the number of moles of base required to neutralize each acidic hydrogen is proportional to the number of acidic hydrogens in the acid.

Let's analyze the acids:

1.
`\(\ce{HNO2}\)`: Monoprotic acid, so one mole of base reacts with one mole of acid.

2.
`\(\ce{HC2H3O2}\)`: Monoprotic acid, similar to
`\(\ce{HNO2}\)`.

3.
`\(\ce{H2SO3}\):` Diprotic acid, so each mole of acid can react with two moles of base.

4.
`\(\ce{H2C6H6O6}\)`: Diprotic acid, similar to
`\(\ce{H2SO3}\)`.

5.
`\(\ce{H3AsO4}\)`: Triprotic acid, so each mole of acid can react with three moles of base.

Considering the molarity and the number of acidic hydrogens, we can rank the acids from least to most volume of base needed:

1.
`\(\ce{HNO2}\)` (monoprotic)

2.
`\(\ce{HC2H3O2}\)` (monoprotic)

3.
`\(\ce{H2SO3}\)` (diprotic)

4.
`\(\ce{H2C6H6O6}\)` (diprotic)

5.
`\(\ce{H3AsO4}\)` (triprotic)

Therefore, the ranking from least to most volume of base needed is
`\(\ce{HNO2} < \ce{HC2H3O2} < \ce{H2SO3} < \ce{H2C6H6O6} < \ce{H3AsO4}\)`.