For H,P - HP- + H* • The weak acid H,P dissociates a little and the even weaker acid HP- much less than that. • Treat the system as a monoprotic acid. • Set up and solve the equation [H*][HP] _ x? [H,P] where [H*] = [HP] = x and [H,P] = F - x. x2 Ka1= 10-2.950 = 0.00112 %3D %3D F-x 0.050-x Solve using the quadratic equation to obtain x = [H*] = 0.0230 M pH = -log[H*] = 1.64 %3D Please show step by step the math that my professor used to get to x=.0230M and why
For H,P - HP- + H* • The weak acid H,P dissociates a little and the even weaker acid HP- much less than that. • Treat the system as a monoprotic acid. • Set up and solve the equation [H*][HP] _ x? [H,P] where [H*] = [HP] = x and [H,P] = F - x. x2 Ka1= 10-2.950 = 0.00112 %3D %3D F-x 0.050-x Solve using the quadratic equation to obtain x = [H*] = 0.0230 M pH = -log[H*] = 1.64 %3D Please show step by step the math that my professor used to get to x=.0230M and why
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter13: Acids And Bases
Section: Chapter Questions
Problem 98QAP: Consider the following six beakers. All have 100 mL of aqueous 0.1 M solutions of the following...
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![For H2P
- HP- + H+
• The weak acid H,P dissociates a little and the
even weaker acid HP- much less than that.
Treat the system as a monoprotic acid.
• Set up and solve the equation
[H*][HP°]
(H,P]
where [H*] = [HP] = x and [H,P] = F - x.
x2
Ka1= 10-2.950 = 0.00112
%3D
%3D
%3D
F-x
0.050-x
Solve using the quadratic equation to obtain
x = [H*] = 0.0230 M
pH = -log[H*] = 1.64
%3D
Please show step by step the math that
my professor used to get to x=.0230M
and why](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5d3979a-8847-4bcd-b900-72b0b330afd3%2Faa3a42e5-997d-42bb-9b60-3562b940d45d%2F3b5wfhf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For H2P
- HP- + H+
• The weak acid H,P dissociates a little and the
even weaker acid HP- much less than that.
Treat the system as a monoprotic acid.
• Set up and solve the equation
[H*][HP°]
(H,P]
where [H*] = [HP] = x and [H,P] = F - x.
x2
Ka1= 10-2.950 = 0.00112
%3D
%3D
%3D
F-x
0.050-x
Solve using the quadratic equation to obtain
x = [H*] = 0.0230 M
pH = -log[H*] = 1.64
%3D
Please show step by step the math that
my professor used to get to x=.0230M
and why
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