buffered solutions

# buffered solutions - [HCO3-] / [H2CO3] = 4.3 x 10-7 /...

This preview shows pages 1–2. Sign up to view the full content.

buffered solutions  - contains weak conjugate acid-base pair   resists drastic pH changes able to neutralize both H+ and OH- ions conjugate acid-base pair can't consume each other usually created by combining a weak acid/base w/ its salt pH of buffer = log ([base] / [acid]) - log (Ka) o pH = pKa + log ([X-] / [HX]) buffer capacity - amount of acid/base the buffer can neutralize before pH begins changing add strong acid >> [HX] increases, [X-] decreases add strong base >> [X-] incresaes, [HX] decreases If the buffer system in blood is made up of carbonic acid and sodium bicarbonate, what is the base to acid ratio if the  pH is measured to be 7.41?   Given: o pH = 7.41 o Ka of carbonic acid = 4.3 x 10-7 o Ka = [H+] [X-] / [HX] [H+] = 10-7.41 4.3 x 10-7 = [H+] [HCO3-] / [H2CO3]

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: [HCO3-] / [H2CO3] = 4.3 x 10-7 / 10-7.41 = 11 How many moles of NaOBr should be added to 1.00 L of 0.050 M HOBr to form a buffer w/ pH 8.80? Given: o pH = 8.80 o Ka HOBr = 2.5 x 10-9 o Ka = [H+] [X-] / [HX] assume that the NaOBr has negligible effect on the volume [H+] = 10-8.8 2.5 x 10-9 = [H+] [NaOBr] / [HOBr] [NaOBr] = (2.5 x 10-9) (0.050) / (10-8.8) = 0.079 In a buffer of acetic acid and sodium acetate where the pH is 5, find the molarity of sodium acetate if the molarity of acetic acid is 0.10 Given: o pH = 5 o [acetic acid] = 0.10 o Ka = 1.8 x 10-5 o Ka = [H+] [X-] / [HX] [H+] = 10-5 1.8 x 10-5 = [H+] [sodium acetate] / [acetic acid] [sodium acetate] = (1.8 x 10-5) (0.1) / (10-5) = 0.18...
View Full Document

## buffered solutions - [HCO3-] / [H2CO3] = 4.3 x 10-7 /...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online