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Acid-Base Equilibria: pH Calculation with Quadratic Equations and Henderson-Hasselbalch, Study notes of Chemistry

Solutions to recitation section problems related to calculating ph of acid-base systems using quadratic equations and henderson-hasselbalch equation. The problems involve chlorous acid, hydronium ions, and methylamine. Students will learn how to set up and solve the equations to determine the ph of the solutions.

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2021/2022

Uploaded on 09/12/2022

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Recitation Section Problems
February 17, 2020
Solutions
1. The pKaof chlorous acid (HClO2) is 1.96. A buffer is made that is 1.20 M in chlorous
acid and 0.50 M in sodium chlorite (NaClO2). Calculate the pH of the resulting solution.
You must solve the full quadratic equation for this system.
Answer:
HClO2(aq)+ H2O(`)*
)H3O+
(aq)+ ClO
2(aq)
Ka=[ClO
2][H3O+]
[HClO2]= 101.96 = 1.1×102
[HClO2] [ClO
2] [H3O+]
initial 1.20 M 0.50 M 0 M
change -y y y
equilibrium (1.20 -y) M (0.50 + y) M yM
y(y+ 0.50)
1.20 y= 1.1×102
y2+ 0.51y0.013 = 0
y=0.51 ±[(0.51)2+ 4(0.013)]1/2
2= 2.5×102negative solution ignored
[H3O+] = 2.5×102M pH = log10(2.5×102) = 1.61
1
pf3
pf4

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Recitation Section Problems

February 17, 2020

Solutions

  1. The pKa of chlorous acid (HClO 2 ) is 1.96. A buffer is made that is 1.20 M in chlorous acid and 0.50 M in sodium chlorite (NaClO 2 ). Calculate the pH of the resulting solution. You must solve the full quadratic equation for this system. Answer: HClO2(aq) + H 2 O(`) ⇀↽ H 3 O+(aq) + ClO− 2(aq)

Ka =

[ClO− 2 ][H 3 O+] [HClO 2 ]

= 10−^1.^96 = 1. 1 × 10 −^2

[HClO 2 ] [ClO− 2 ] [H 3 O+] initial 1.20 M 0.50 M 0 M change - y y y equilibrium (1.20 -y) M (0.50 + y) M y M

y(y + 0.50)

  1. 20 − y

= 1. 1 × 10 −^2

y^2 + 0. 51 y − 0 .013 = 0

y =

− 0. 51 ± [(0.51)^2 + 4(0.013)]^1 /^2

= 2. 5 × 10 −^2 negative solution ignored

[H 3 O+] = 2. 5 × 10 −^2 M pH = − log 10 (2. 5 × 10 −^2 ) = 1. 61

  1. If 0.10 moles of hydronium ions are added to 1.00 L of the solution in the previous problem, calculate the final pH. Answer:

nHClO 2 nClO− 2 nH 3 O+ initial 1.20 0.50 0 M change 0.10 -0.10 y equilibrium 1.30 M 0.40 M y M

y(0.40)

  1. 30

= 1. 1 × 10 −^2

y = 1. 1 × 10 −^2

(

  1. 30
  2. 40

) = 3. 6 × 10 −^2

pH = − log 10 (3. 6 × 10 −^2 ) = 1. 45

  1. Calculate the pH of a solution that is made by combining 0.100 L of the buffer from problem 3 with 0.0100 L of 0.100 M sodium hydroxide. Answer: nCH 3 N H 2 = (0.25 mol L−^1 )(0.100 L) = 0.025 mol nCH 3 N H 3 + = (0.35 mol L−^1 )(0.100 L) = 0.35 mol

nOH−^ added = (0.100 mol L−^1 )(0.0100 L) = 0.0010 mol

pH = pKa + log 10

[CH 3 NH 2 ]

[CH 3 NH+ 3 ]

= 3.36 + log 10