Post Lecture Homework Chapter 16 t pH Changes in Buffers t pH Changes in Buffers
ID: 509150 • Letter: P
Question
Post Lecture Homework Chapter 16 t pH Changes in Buffers t pH Changes in Buffers Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 608 mol of NaA When a solution contains a weak acid and its in 2.00 Lof solution? The dissociation constant Ka of HA is 5.66 x 10 conjugate base or a weak base and its conjugate acid, it will be a buller solution. Buffers resist Express the pH numerically to three decimal places. change in pH following the addition of acid or base. A buffer solution prepared fram a weak acid (HA) and its conjugate base (A) is represented as pH 6,123 HA(aq) H (aq) A (aq) Submit Hints My Answers Give Up Review Part The buffer will follow Le Chateliers principle. If acid is added, the reaction shifts to consume the added Correct H forming more HA. When base is added, the base will react with H+, reducing its concentration. Since both the acid and base exist in the same volume, we can skip the conoentration caloulations and use the number of males in the Henderson-Hasselbalch equation to calculate the pH. The The reaction then shifts to replace H through the dissociation ofHA into H and A .In both answer will be the same. instances, H+] tends to remain constant. significant Figures Feedback: Your answer 6.24 was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later The pH of a buffer is calculated by using the calculation in this item, keep all the digits and round as the final step before submitting your Henderson-Hasselbalch equation: pH pKa+ log HA Part B What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume chang on the addition of the acid. Express the pH numerically to three decimal places.Explanation / Answer
PART B)
moles of HA = 0.809
moles of NaA = 0.608
pKa = 6.25
moles of HCl added = 0.150 mol
pH = pKa + log [salt - C / acid + C]
= 6.25 + log [0.608 - 0.150 / 0.809 + 0.150]
= 5.929
pH = 5.929
PART C)
moles of NaOH added = 0.195 mol
pH = pKa + log [salt + C / acid - C]
= 6.25 + log [0.608 + 0.150 / 0.809 - 0.195]
= 6.34
pH = 6.342