Question

Determine the volume (in mL) of 0.411 M hydrochloric acid (HCl) that must be added to 448 mL of 0.914 M sodium butanoate (NaC3H7COO) to yield a pH of 5.48. Assume the 5% approximation is valid and report your answer to 3 significant figures. A table of pKa values can be found here. pKa=4.82

Answer 1

Let us say that the volume of HCl must be added = X mL

Moles of HCl = 0.411X mmol

Moles of sodium butanoate = 448 x 0.914 = 409.472 mmol

0.411X mmol of HCl will react with 0.411X mmol of sodium butanoate to form 0.411X mmol of butanoic acid

Therefore remaining sodium butanoate = (409.472 - 0.411X) mmol

Total volume of the solution = (448 + X) mL

Molarity of butanoic acid formed = 0.411X /(448 + X) M

Molarity of remaining sodium butanoate = (409.472 - 0.411X)/(448 + X) M

Given, pKa of butanoic acid = 4.82

pH of the solution = 5.48

Now, from Henderson-Hasselbalch equation,

pH = pKa + log[base]/[acid]

or, 5.48 = 4.82 + log(409.472 - 0.411X)/0.411X      (Cancelling (448 + X) from numerator and denominator)

or, log(409.472 - 0.411X)/0.411X = 0.66

or, (409.472 - 0.411X)/0.411X = 10^0.66 = 4.57

or, (409.472 - 0.411X) = 1.88X

or, 2.291X = 409.472

or, X = 179 mL

Therefore, the volume of HCl that must be added = 179 mL