What is pH of the buffer solution?
A buffer, consisting of H2PO4- and HPO42-, helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the pH of a soft drink in which the major buffer ingredients are 6.90 g of NaH2PO4 and 4.50 g of Na2HPO4 per 355 mL of solution?
Humberto, you are right, just don't round so much.
The actual answer is 6.949, which is a significant difference from your answer.
- 1 decade agoFavorite Answer
This should be done in this way. The two salts that you are adding to make the buffer will produce H2PO4- and HPO42-. The main equilibrium reaction will be
H2PO4- == H+ + HPO42- ka2 = 6.32e-8
ka2 = [H+][HPO42-]/[H2PO4-]
-log(ka2) = -log[H+] - log([HPO42-]/[H2PO4-])
pka2 = pH - log([HPO42-]/[H2PO4-])
pH = pka2 + log([HPO4-]/[H2PO42-])
# of moles of H2PO4- = 6.9 / (22.99 + 2 + 30.974 + 16*4)
= 0.05752 mole
[H2PO4-] = 0.05752/0.355 = 0.162 M
# of moles of HPO42- = 4.5 / (22.99*2 + 1 + 30.97 + 16*4)
= 0.0317 mole
[HPO42-] = 0.0317 / 0.355 = 0.0893 M
pH = -log(6.32e-8) + log(0.0893/0.163)
pH = 6.941
Please note that the ka used in this case is for 25 deg C, and is obtained from an analytical chemistry text book, which should be more accurate compared to a general chemistry text book.
Note that the previous person's answer is very off. Remember pH is in log scale. For a pH different by 0.5 is huge in terms of H+ concentration.
- pirrieLv 43 years ago
once you've 10 ml of the buffer, then you truthfully have 0.002 moles of lactic acid and nil.003 moles of sodium lactate present. once you upload 10 ml of 0.01 M HCl, you're including 0.0001 moles of H+, which will react with the lactate to furnish lactic acid. you'll now have 0.0021 moles lactic acid and nil.0029 moles lactate, yet your quantity has now higher to twenty mL. nevertheless, considering they're interior a similar answer, you should use the kind of moles contained in the Henderson-Hasselbach equation contained in the ratio of [base]/[acid] to verify pH. pH = pKa + log [base]/[acid] = 3.80 5 + log 0.0029/0.0021 = 3.ninety 9 I extremely suspect that the pH of the buffer you calculated earlier to the addition of HCl is inaccurate, or the concentrations are incorrect. If pKa is 3.80 5, your pH can't be below that if the most objective of the bottom of the conjugate pair is larger than the acid.
- 1 decade ago
first, find the molar concentration of H2PO4. Once you work out the arithmetic, it turns out to be .2
the molar concentration of HPO4 turns out to be .132
with a Ka of 6.2x10^(-8) you just plug and chug.
GIVE ME A BREAK IM FREAKING DRUNK