Question

Part A A beaker with 135 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.30 mL of a 0.450 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Part B As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the potential to kill weeds effectively. A sparingly soluble salt, Herbigon is dissolved in 1 M acetic acid for technical reasons having to do with its production. You have determined that the solubility product Ksp of Herbigon is 9.40×10−6. Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH3COO, where "X" represents the top-secret cation of the salt). It is this cation that kills weeds. Since it is critical to have Herbigon dissolved (it won't kill weeds as a suspension), you are working on adjusting the pH so Herbigon will be soluble at the concentration needed to kill weeds. What pH must the solution have to yield a solution in which the concentration of X+ is 1.50×10−3 M ? The pKa of acetic acid is 4.76.

Answer 1

part A)

millimoles of acid + conjugate base = 135 x 0.1 =13.5

pH = pKa + log [conjugate base / acid ]

5.00 = 4.74 + log [conjugate base / acid ]

1.82 = conjugate base / acid

conjugate base = 1.82 acid -------------> 2

from 1 and 2

1.82 acid + acid = 13.5

acid = 4.79

conjugate base = 8.71

millimoles of strong acid = C = 8.30 x 0.450 = 3.74

on additon of C millimoles of acid to buffer

pH = pKa + log [conjugate base - C / acid + C]

pH = 4.74 + log (8.71 - 3.74 / 4.79 + 3.74)

pH = 4.506

pH chnage = 4.506 - 5.000

= -0.49

pH chnage  = - 0.49

part B)

salt concentration = C = 1.50 x 10^-3 M

pKa = 4.76

salt if from weak acid and strong base . so pH > 7

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/2 [4.76 + log 1.50 x 10^-3 ]

pH = 7.97