Question

A beaker with 1.10×10² 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 4.90 mLof a 0.380 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.

ΔpH=

Answer 1

The reaction involves in acetic acid buffer is as follows.

Let there are "x" M acetic acid and "y" M acetate in the solution.x + y = 0.1 , therefore, y = 0.1-x

                   CH₃COOH (Aq) + H₂O     <---------> CH₃COO^- (aq) + H⁺ (aq)

pH of the solution = 5.0

Since -   pH = pKa + log [AcONa] / [ AcOH]

              5 = 4.74 + log [(0.1-x) / (0.1)]

Hence - [(0.1-x) / x] = 1.82

               x = 0.035 M

Therefore, y = 0.1 - x = 0.1 - 0.035 = 0.065 M

So, at equilibrium [AcOH] = 0.035 M and [AcONa] = 0.065 M

Moles_CH3COOH = (0.035 M*0.110 L) = 0.00385 mol

Moles_CH3COONa = (0.065 M*0.110 L) = 0.00715 mol

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When 4.90 mLof a 0.380 M HCl (0.001862 mol HCl) is added to the above solution, equilibrium shifts towards left.

Thus, Number of mols of AcONa left = 0.00715 mol - 0.001862 mol = 0.005288 mol

Number of mols AcOH formed = 0.00385 mol + 0.001862 mol = 0.005712 mol

Hence, pH = 4.74 + log (0.005288 / 0.005712)   [Volume is constant]

                  = 4.70

Hence pH change during this addition of HCl = 5- 4.70 = 0.30