Calculate the number of moles of HCl (g) that must be added to 1.0L of 1.0M NaC2H302 to produce a solution buffered at each pH. (Ka for HC2H3O2 = 1.8X10^-5)a) pH=pKab) pH=3.92c) pH=5.49
Question
Calculate the number of moles of HCl (g) that must be added to 1.0L of 1.0M NaC2H302 to produce a solution buffered at each pH. (Ka for HC2H3O2 = 1.8X10^-5)a) pH=pKab) pH=3.92c) pH=5.49
Solution
To solve this problem, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). Here, [A-] is the concentration of the base (sodium acetate, NaC2H3O2) and [HA] is the concentration of the acid (acetic acid, HC2H3O2).
a) When pH = pKa, the ratio of [A-]/[HA] is 1. This means that the concentration of the acid is equal to the concentration of the base. Since the initial concentration of the base is 1.0 M, we need to add 1.0 mole of HCl to react with the base and form 1.0 mole of the acid.
b) When pH = 3.92, we first need to calculate the pKa from the Ka. pKa = -log(Ka) = -log(1.8x10^-5) = 4.74. Then we substitute the values into the Henderson-Hasselbalch equation: 3.92 = 4.74 + log([A-]/[HA]). Solving for [HA], we get [HA] = [A-] * 10^(4.74 - 3.92) = 1.0 M * 10^0.82 = 6.6 M. Since the initial concentration of the base is 1.0 M, we need to add 6.6 moles of HCl to form 6.6 moles of the acid.
c) When pH = 5.49, we substitute the values into the Henderson-Hasselbalch equation: 5.49 = 4.74 + log([A-]/[HA]). Solving for [HA], we get [HA] = [A-] * 10^(4.74 - 5.49) = 1.0 M * 10^-0.75 = 0.18 M. Since the initial concentration of the base is 1.0 M, we need to add 0.18 moles of HCl to form 0.18 moles of the acid.
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