Question

Solution C is a 1.00 L buffer solution that is 1.420 M in acetic acid and 0.67 M in sodium acetate. Acetic acid has a pKa of 4.74. What is the pH change of this solution upon addition of 0.10100 mol of HCl? Enter a negative number to 3 decimal places.

Answer 1

Answer:

The correct answer is 0.10.

Explanation:

Based on the given question, in a buffer solution of 1 liter, the molarity of acetic acid is 1.420 M and the molarity of sodium acetate is 0.67. The pKa value of acetic acid given is 4.74, now the pH of buffer is,  

pH of buffer = pKa + log ([CH3COONa]/[CH3COOH])

= 4.74 + log (0.67/1.420)

= 4.74 + (-0.326)

= 4.41  

Now 0.10100 mol of HCl is added, the HCl reacts with sodium acetate to give,  

CH3COONa + HCl = CH3COOH + NaCl

Now the concentration of CH3COONa becomes = 0.67-0.101 = 0.57 M, and the new concentration of CH3COOH becomes = 1.420 + 0.101 = 1.52 M

Now the new pH will be,  

= pKa + log (0.57/1.52)

= 4.74 + (-0.426)

= 4.31

The pH change is 4.41-4.31 = 0.10