What is the pH of this buffer?
a buffer solution is prepared by mixing .500 L of 0.10 M NaOCl and 0.500 L fo 0.20 M HOCl. What is the pH of this buffer? (Ka(HOCl)= 3.2 x 10^-8)
Answers:
Moles of NaOCl = (0.500 L) * (0.10 M) = 0.0500 moles
Moles of HOCl = (0.500 L) * (0.20 M) = 0.100 moles
Total volume of solution = 0.500 L + 0.500 L = 1.000 L
Concentration (molarity) of new NaOCl solution = 0.0500 moles / 1.000 L = 0.0500 M.
Concentraion of new HOCl solution = 0.100 moles / 1.000 L = 0.100 M.
Now, use the henderson-hasselbach equation:
pH = pKa + log ([base]/[acid]), or in this suitcase,
pH = pKa + log ([NaOCl]/[HOCl])
pKa refers to pKa of the acid in question. The acerbic in this case is HOCl.
So pKa = -log (Ka) = -log (3.2 x 10^-8) = 7.49
Now just plug and chug what you know.
pH = pKa + log ([NaOCl]/[HOCl])
pH = 7.49 + log (0.0500 M / 0.100 M) = 7.19. There's your answer. Source(s): UW pharmacy student
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Answers:
Moles of NaOCl = (0.500 L) * (0.10 M) = 0.0500 moles
Moles of HOCl = (0.500 L) * (0.20 M) = 0.100 moles
Total volume of solution = 0.500 L + 0.500 L = 1.000 L
Concentration (molarity) of new NaOCl solution = 0.0500 moles / 1.000 L = 0.0500 M.
Concentraion of new HOCl solution = 0.100 moles / 1.000 L = 0.100 M.
Now, use the henderson-hasselbach equation:
pH = pKa + log ([base]/[acid]), or in this suitcase,
pH = pKa + log ([NaOCl]/[HOCl])
pKa refers to pKa of the acid in question. The acerbic in this case is HOCl.
So pKa = -log (Ka) = -log (3.2 x 10^-8) = 7.49
Now just plug and chug what you know.
pH = pKa + log ([NaOCl]/[HOCl])
pH = 7.49 + log (0.0500 M / 0.100 M) = 7.19. There's your answer. Source(s): UW pharmacy student
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