Calculate the mass in g of aniline (C6H5NH2, Kb = 4.3 x 10-10) that must be added to 1.4 x 102 mL of 0.24 M (C6H5NH3)Cl solution in order to generate a buffer solution of pH 5.56. Give your answer to 2 significant figures.
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Use the Henderson-Hasselbalch equation.
Aniline is the base and aniline chloride is the acid.
pH = pKa + log [(base)/(acid)]
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To calculate the mass of aniline (C6H5NH2), we need to use the Henderson-Hasselbalch equation and the concept of buffer solutions.
1. Start by writing down the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
2. Determine the pKa value:
The pKa is the negative logarithm of the acid dissociation constant (Ka) for aniline. Given that Kb = 4.3 x 10^(-10), we can calculate Ka using the relation: Ka x Kb = Kw (water dissociation constant).
Kw = 1.0 x 10^(-14) at 25 degrees Celsius.
Therefore, Ka = Kw/Kb = (1.0 x 10^(-14))/(4.3 x 10^(-10)) = 2.3 x 10^(-5)
Take the negative logarithm of Ka to find pKa: pKa = -log(2.3 x 10^(-5)) = 4.64
3. Substitute the given information into the Henderson-Hasselbalch equation:
pH = 5.56 (given)
pKa = 4.64 (calculated from step 2)
[A-]/[HA] = (C6H5NH2)/(C6H5NH3Cl)
4. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
5. Calculate the ratio of [A-]/[HA]:
[A-]/[HA] = 10^(5.56 - 4.64) = 7.59
6. Use the ratio of [A-]/[HA] to find the concentration of C6H5NH2:
C6H5NH2 = (7.59)(0.24 M) = 1.82 M
7. Calculate the moles of C6H5NH2:
Moles = concentration × volume
Moles = 1.82 M × 0.102 L = 0.185 g
Therefore, the mass of aniline (C6H5NH2) that must be added is approximately 0.185 grams to generate a buffer solution of pH 5.56.