Question

Calculate the pH of a buffer made from mixing 7.3 mL of 0.229 M NaC2H3O2 and 12.2 mL of 0.458 M HC2H3O2. The Ka of HC2H3O2 is 1.8 x 10^-5.

Answer 1

Answer:

pH = 4.22

Explanation:

NalgH3 2 First we need to find the number of moles of and HC₂ H₂O₂.

Calculation of number of moles of NaC₂H₂O2" Volume = 7.3 mL egiven) Concentration = 0.229 M = 0.229 mol (given) No. of moles = 7.3 mx x 0.229 mol No. of moles = 1.6717 mmol No. of moles of NaCHP2 Calculation of number of moles of HC₂H₂ 2: volume = 12. 2mL (given) Concentreation = 0.458 M = 0.458 mal egiven) No. of moles = 12.2 ml x 0.458 mol TNo. of moles=5.5876 mmol No. of moles of H₂C2H₂O

Now we need to calculate pKa from the given value of Ka,

Calculation of pka! Ka= 1.88 10 5 (given) pka = -dog (ka) pka = -log (1.8x103) [pka = 4.74

this equation is also known as Henderson Hasselbalch equation.

now, we can use the following equation to calendate the pH of this solution. /No. of males of base) pH = pka + log No. of males of acid pH = pka + 80g (n. 4 maler og base

In this solution, HC2H₂O2 is the acid. Na C₂ H₂O2 is the conjugate base.

e No. of moles of base = 1. 6717 mmol No. of moles of acid = 5.587 6 mmol pka = 4.74 Put these values in the equation fore pH, pH = 4.74 + log (1.6717 minnel) 5.587 6 minol) pH = 4.74 + dog (0.2992) pH = 4.74+ (-0.52) JPH = 4.22