A buffer solution is prepared by mixing 22.0 mL of 0.384 M benzoic acid with 73.
ID: 925540 • Letter: A
Question
A buffer solution is prepared by mixing 22.0 mL of 0.384 M benzoic acid with 73.2 mL of 0.0864 M sodium benzoate. A table of pKa values can be found here. Calculate the pH (to two decimal places) of this solution. Assume the 5% approximation is valid and that the volumes are additive. Calculate the pH (to two decimal places) of the buffer solution after the addition of 509 mL of a 0.000430 M solution of rubidium hydroxide to the existing buffer solution. Assume the 5% approximation is valid and that the volumes are additive.Explanation / Answer
Apply buffer equation
pH = pKa + log(A-HA)
pKa= 4.20
A- = M*V/(V1+V2) = 0.384*22/(22+73.2) =0.08873
HA = MV/(V1+V2) = 73.2*0.0864/(22+73.2) =0.0664
substitute
pH = pKa + log(A-HA) = 4.20 + log(0.08873/0.0664) = 4.3259
pH = 4.3259
2)
If
V = 509 ml of M = 0.00043 RbOH
mmol of base = M*V = 0.00043*509 = 0.21887 mmol
mmol of acid = M*V = 0.0664*(95.2) = 6.321
mmol of conjugate= MV = 0.08873*(95.2) = 8.4470
after addition
mmol of acid = 6.321 - 0.21887 = 6.10213
mmol of conjugate = 8.4470 +0.21887 = 8.66587
pH = pKa + log(A-HA) = 4.20 +loG(8.66587/6.10213) = 4.352
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