onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re
1 answer:
Answer:
The volume of the final solution, V = 0.0305L
Explanation:
Number of moles = Concentration * volume
Concentration of HA = 1.00 * 10⁻⁴M
Volume of HA = 1000mL = 1 L
Number of moles of HA = 1.00 * 10⁻⁴ * 1
Number of moles of HA = 1.00 * 10⁻⁴ mols
Equation of reaction:
HA → H⁺ + A⁻
If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.
Since only 16% dissociation occurs = 0.16
Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols
Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols
Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left
Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴
Number of mols of HA left = 8.4 * 10⁻⁵mols
Concentration = num of moles/volume
Let the volume of the final solution be V
Conc of HA = 8.4 * 10⁻⁵/V
Conc of H⁺ = 1.6 * 10⁻⁵/V
Conc of A⁻ = 1.6 * 10⁻⁵/V
To calculate the dissociation constant
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Answer:
400 J
Explanation:
The efficiency of a machine is defined as
where
is the work output
is the work input
For this machine, we know the following
is the work in input
is the efficiency
Solving for the work output, we find