Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid
ID: 975233 • Letter: P
Question
Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.609 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66 × 10-7 Express the pH numerically to three decimal places. | pH= 6.124 Express the ph numericaily to three decimal places. Submit Hints My Answers Give Up Review Part Correct Since both the acid and base exist in the same volume, we can skip the concentration calculations and use the number of moles in the Henderson-Hasselbalch equation to calculate the pH. The answer will be the same.Explanation / Answer
B) When 0.15 moles of HCl is added,
HCl + NaA ----------> HA + NaCl
That means additional amount HA is produced.
Hence,
pH = -logKa + log [NaA] -[HCl] / [HA]+[HCl]
= -log (5.66x 10-7) + log [0.609-0.15 ]/[0.809+0.15]
= 5.93
pH = 5.93
C)
When 0.195 moles of NaOH is added,
NaOH+ HA ----------> NaA + H2O
That means additional amount NaA is produced.
Hence,
pH = -logKa + log [NaA] +[NaOH] / [HA]-[NaOH]
= -log (5.66x 10-7) + log [0.609+0.195 ]/[0.809-0.195]
= 6.36
pH = 6.36