The following equilibrium constants will be useful foe some of the problems. Ref
ID: 491774 • Letter: T
Question
The following equilibrium constants will be useful foe some of the problems. Refer to Equilibrium Constants. What is the pH of a solution that is 0.20 M in HCN and 0.15 M in KCN? a. 7.60 b. 8.40 c. 8.80 d. 9.27 e. 10.10 Refer to Equilibrium Constants. Which one of the following combinations cannot produce a buffer solution? a. HNO_3 and NaNO_2 b. HCN and NaCN c. HCIO, and NaCIO_4 d. NH_3 and (NH_4)_2SO_4 e. NH_3 and NH_4Br Refer to Equilibrium Constants. If 100. mL of 0.040 M NaOH solution is added to 100. mL of solution which is 0.10 M in CH_3COOH and 0.10 M in NaCH_3COO, what be will the pH of the new solution ? a. 4.74 b. 4.81 c. 489 d. 5.00 e. 5.11 Refer to Equilibrium Constants. What is the Delta pH in a solution which is 0.10 M in sodium acetate and 0.20 M in acetic acid, after 0.05 mol/L of hydrochloric acid has been bubbled? a. -0.40 b. +0 60 c. -0.08 d. +0.30 e. -0.18Explanation / Answer
21. This is a buffer solution : weak acid HCN and its conjugate base KCN
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
pKa = -log(4 x 10^-10) = 9.4
So,
pH = 9.4 + log(0.15/0.2) = 9.27
Answer : d. 9.27
22. A buffer solution is a combination of weak acid-conjugate base or weak base-conjugate acid
Among the given choices, HClO4 is a strong acid and hence cannot form a buffer solution.
Answer : c. HClO4 and NaClO4
23. Initial pH of the buffer system,
pH = pKa + log(base/acid)
= 4.75 + log(0.1/0.1) = 4.75
when NaOH = 0.04 M x 100 ml = 4 mmol was added
new pH = 4.75 + log[(0.1 M x 100 ml + 4 mmol)/(0.1 M x 100 ml - 4 mmol)]
= 5.11
Answer : e. 5.11
24. For the buffer system,
initial pH = pKa + log(base/acid)
= 4.75 + log(0.10/0.20) = 4.45
When HCl = 0.05 mol/L was added
new pH = 4.75 + log[(0.1 - 0.05)/(0.2+ 0.05)]
= 4.05
thus,dpH = 4.05 - 4.45 = -0.40
Answer : a. -0.40
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.