Question

A beaker with 2.00×102 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 7.20 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.

Answer 1

Solution :-

pH= 5.00

total molarity = 0.100 M

Volume = 200 ml

Volume of HCl added = 7.20 ml

Molarity of HCl = 0.450 M

Change in the pH= ?

Lets find the ratio of the acid and conjugate base at initial pH

PH= pka + log [base]/[acid]

X+y = 0.100

5.00 = 4.74 + log [base]/[acid]

5.00 – 4.74 = log [y]/[x]

0.26 = log [y/x]

Antilog [0.26] = [y/x]

1.82 = [y/x]

1.82x= y

Total molarity = 0.100

X+ [1.82x] =0.100

2.82x= 0.100

X= 0.100/2.82

X= 0.03546 M acid molarity

Molarity of conjugate base = 0.100 M – 0.03546 M = 0.06454 M

Volume is 200 ml = 0.200 L

Lets calculate the moles of acid and conjugate base

Moles of acid = 0.03546 mol perL * 0.200 L = 0.007092 mol

Moles of acetate = 0.06454 mol per L * 0.200 L = 0.01291 mol

Moles of HCl = 0.450 mol per L * 0.00720 L = 0.00324 mol

New moles of acid and base after adding HCl

Moles of acid = 0.007092 mol + 0.00324 mol = 0.01033 mol

Moles of acetate = 0.01291 mol – 0.00324 mol = 0.00967 mol

Total volume of solution after adding HCl = 200 ml + 7.20 ml = 207.2 ml = 0.2072 L

New molarity of the acid and base at total volume are as follows

[acid] = 0.01033 mol / 0.2072 L = 0.0499 M

[conjugate base] = 0.00967 mol / 0.2072 L = 0.0467 M

Now lets calculate the pH

pH = pka + log ([base]/[acid])

pH= 4.740 + log [0.0467 /0.0499]

pH= 4.97

change in pH= 4.97 – 5.00 =   -0.030

so the change in the pH is -0.030