Question

imagine mixing 25.0 ml of 0.10 M acetic acid (Ka=1.8*10^-5) with 25.0 ml of 0.10 M sodium acetate and determine a) The initial pH of the buffer b) pH of 20.0 ml sample of buffer with 1.0 ml of 0.10M Hal added c) pH of another 20 mL sample of buffer with 1.0 ml of 0.10 M MaOH added

Answer 1

acid = acetic acid                   salt = sodium acetate

K_a = 1.8 x 10^-5

P^Ka = - log K_a = - log (1.8 x 10^-5) = 5 - log 1.8 = 4.74

a)

According to Henderson–Hasselbalch equation

[acid] = (0.10 x 25)/ (25+25) = 0.05 M

[salt] =( 0.10 x25)/ (25+25) = 0.05 M

P^H = P^Ka + log [salt]/[acid] = 4.74 + log (0.05/0.05) = 4.74 + 0 = 4.74

b)

moles of acid and salt present in 20 ml of buffer

acid moles = molarity x volume in lit = 0.05 x 0.02 = 0.001 moles

salt moles = molarity x volume in lit = 0.05 x 0.02 = 0.001 moles

number of moles in 1 ml of 0.1M of HCl = molarity x volume in lit = 0.1 x 0.001 = 0.0001 moles

acid moles after adding 1.0 ml of 0.10M HCl = 0.001 +0.0001 = 0.0011

P^H = P^Ka + log [salt]/[acid] = 4.74 + log (0.001/0.0011) = 4.74 -0.04= 4.70

c)

CH₃COOH + NaOH ------> CH₃COONa + H₂O

number of moles in 1 ml of 0.1M of NaOH = molarity x volume in lit = 0.1 x 0.001 = 0.0001 moles

salt moles after adding 1.0 ml of 0.10M NaOH = 0.001 +0.0001 = 0.0011

acid moles after adding 1.0 ml of 0.10M NaOH = 0.001 -0.0001 = 0.0009

P^H = P^Ka + log [salt]/[acid] = 4.74 + log (0.0011/0.0009) = 4.74 + 0.09= 4.83