Monday
March 30, 2015

Homework Help: Chemistry

Posted by Vanessa on Monday, March 22, 2010 at 7:26pm.

Wilh this form a buffer?
50 mL of .10 M HCl; 35 mL of .150 M NaOH

WORK:
Moles of HCl= .005
Moles of NaOH= .00525

total litres of solution= .05 + .035= .085 L

[HCl]= .005 mol/.085 L = .0588 M
[NaOH]= .00525 mol/ .085 L = .0618 M

Ka of HCl= 1.3*10^6
pKa= -log(1.3*10^6) = -6.114

Plug into H-H equation...
pH= pKa + log [A-]/[HA]
pH= (-6.114)+ log (.0618/.0588)
pH= -6.09???
I am stuck.. I don't think this value is supposed to be negative. How does the pH of the solution tell someone whether or not the soltion is a buffer?

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