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

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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
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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