Macmillan Learning A volume of 500.0 mL of 0.110 M NaOH is added to 585 mL of 0.250 M weak acid (K, = 8.02 x 10-5). What is the pH of the resulting buffer? HA(aq) + OH(aq) → H₂O(1) + A (aq) pH = x10 TOOLS

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Question 24 of 35 - Module 7 Ho X
pH =
x10
TOOLS
Topic: Week 4 Discussion
h
Resources
144
X
A volume of 500.0 mL of 0.110 M NaOH is added to 585 mL of 0.250 M weak acid (K, = 8.02 x 10-5). What is the pH of the
resulting buffer?
HA(aq) + OH(aq) → H₂O(1) + A (aq)
+
Give Up?
DDI
Q
Hint
☐
Check Answer
Attempt 3
12:51 PM
9/29/2022
O
Y
Transcribed Image Text:of 36 - Module 8: Ho X om/sac/7086543#/7086543/23/1 e: > 91.7% Question 24 of 35 - Module 7 Ho X pH = x10 TOOLS Topic: Week 4 Discussion h Resources 144 X A volume of 500.0 mL of 0.110 M NaOH is added to 585 mL of 0.250 M weak acid (K, = 8.02 x 10-5). What is the pH of the resulting buffer? HA(aq) + OH(aq) → H₂O(1) + A (aq) + Give Up? DDI Q Hint ☐ Check Answer Attempt 3 12:51 PM 9/29/2022 O Y
Expert Solution
Step 1

According to Henderson's equation

  PH = PKa + logconjugate baseweak acid

Where within [ ], there are concentration terms.

and PKa = -log(Ka)

 

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