Question

Part A A beaker with 165 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 mol L -1. A student adds 6.70 ml of a 0.320 mol L + HCl solution to the beaker. How much will the pH change? The pK, of acetic acid is 4.780. Express your answer numerically to two decimal places. Use a minus ( - ) sign if the pH has decreased. View Available Hint(s) EVO AXCP Om ? Aph. 1 - 2.6 Submit Previous Answers

Answer 1

According to Henderson-Hasselbalch equation

pH = pKa +log ([Conjugate Base]/[Weak Acid])

5.0 = 4.76 + log ([CH3COO-]/[ CH3COOH])

log ([CH3COO-]/[ CH3COOH]) = 5.0-4.76 = 0.24

([CH3COO-]/[ CH3COOH]) =    100.24 = 1.74

[CH3COO-] = 1.74 [CH3COOH]

Now, because both acetic acid and its conjugate base share the same volume 165 mL, this can be rewritten as

n [CH3COO-] / 0.165 L = 1.74 x n [CH3COOH] / 0.165 L

As volume is same both gets cancelled

Therefore,             n [CH3COO-] = 1.74 x n [CH3COOH]

The total molarity of the acid and conjugate base in this buffer is 0.1 M

[CH3COO-] + [CH3COOH] = 0.1 M

n [CH3COO-] / 0.165 L + n [CH3COOH] / 0.165 L = 0.1 M

n [CH3COO-] + n [CH3COOH] = 0.1 x 0.165 = 0.0165 and ,           

n [CH3COO-] = 1.74 x n [CH3COOH]

From the above two equations,

1.74 x n [CH3COOH] + n [CH3COOH] = 0.0165

n [CH3COOH] (1.74 +1) = 0.0165

n [CH3COOH] = 0.0165/2.74 = 0.0060 moles CH3COOH

[CH3COO-] = 1.74 x 0.0060 = 0.0104 moles of CH3COO-

Now, HCl is added to the above solution

From molarity formula we need to find the amount of HCl

Molarity = No. of moles of solute / Volume of solution in litres

No. of moles of HCl = 0.32 mol/L x 6.7 x 10-3 L = 2.144 x10-3 M = 0.0021moles HCl

After adding HCl, it will be completely consumed in the reaction

n [CH3COO-] = 0.0104 – 0.0021 = 0.0083 moles CH3COO-

n [CH3COOH] = 0.0060 + 0.0021 = 0.0081 moles CH3COOH

Total volume of solution after adding HCl is 165 mL + 6.7 mL = 171.6 mL

Concentration of [CH3COO-] and [CH3COOH] will be

[CH3COO-] = 0.0083 / 0.1716 = 0.0483 M

[CH3COOH] = 0.0081 / 0.1716 = 0.0472 M

From Henderson-Hasselbalch equation the new pH of the solution is

pH = pKa +log ([Base]/[Acid])

pH = 4.76 + log (0.0483/0.0472) = 4.76 + log (1.0233)

pH = 4.76 + 0.0100 = 4.77

Thererfore, change in pH of solution is ∆pH = 4.77-5.0 = - 0.23

Answer is -0.23

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