Which of the following acids (listed with K_a values and would form a buffer wit
ID: 513398 • Letter: W
Question
Explanation / Answer
Question 14
Moles of HCl used = 15 x 0.1 /1000 = 0.0015 Moles
Moles of NaOH Used = 40 x 0.1 /1000 = 0.004 Moles
NaOH is excess. hence NaOH solution only will be rest
Moles NaOH after neutralization = 0.004-0.0015 = 0.0025 Moles
Total Volume = 15 +40 = 55 ml
Concecntration of NaOH = 0.0025 x 1000 / 55 = 0.04545 M
pOH = - log [OH-]
p OH = -log 0.04545 = 1.34
pH + pOH = 14
pH = 14- pOH
pH = 14-1.34 = 12.66
Answer is Option C
Question 15
Moles of HCl used = 17 x 0.2 /1000 = 0.0034 Moles
Moles of NaOH Used = 40 x 0.2 /1000 = 0.008 Moles
NaOH is excess. hence NaOH solution only will be rest
Moles NaOH after neutralization = 0.008-0.0034 = 0.0046 Moles
Total Volume = 17 +40 = 57 ml
Concecntration of NaOH = 0.0046 x 1000 / 57 = 0.081 M
pOH = - log [OH-]
p OH = -log 0.04545 = 1.09
pH + pOH = 14
pH = 14- pOH
pH = 14-1.09 = 12.91
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.