Question

In: Chemistry

A liter of 0.050M pH 7.10 potassium phosphate buffer is needed for a certain experiment.

1)         A liter of 0.050M pH 7.10 potassium phosphate buffer is needed for a certain experiment.

            The following four methods were proposed for the preparation of the buffer. Using the pKa

            value given in your textbook, which of these methods should be used and what is wrong

            with the other proposals? Letter of method which should be used: __________________________

(A)      A 0.050M sample of K2HPO4 was dissolved in about 800.0mL of water, and the pH was

                        adjusted to 7.10 using a standardize pH meter and a 1.00M HCl solution, followed by

                        addition of water to a volume of 1.00L.

(B)       A 0.050M sample of K2HPO4 was dissolved in about 800.0mL of water, and the pH was

                        adjusted to 7.10 using a standardize pH meter and a 1.00M H3PO4 solution, followed by

                        addition of water to a volume of 1.00L.

(C)       Use the Henderson-Hasselbalch equation and 0.050moles = “X” moles of KH2PO4 and

                        “Y” moles of K2HPO4 to calculate the moles of the two salts, weigh out the salts,

                        dissolve, and dilute to 1.00L.

(D)      A 0.050M sample of KH2PO4 was dissolved in about 800.0mL of water, and the pH was

                        adjusted to 7.10 using a standardize pH meter and a 1.00M KOH solution, followed by

                        addition of water to a volume of 1.00L.

Solutions

Expert Solution

The buffer action of potassium phosphate buffer is due to inter conversion of the acid  H2PO4- to the salt HPO42-.

Hence a potassium phosphate buffer contains KH2PO4 and K2HP4.

H2PO4-(aq) + H2O(l) ------> HPO42- (aq) + H3O+  , pKa = 7.21

(A) Here we are taking only K2HPO4 and HCl. Since both are acids we are unable to get the desired concentration of KH2PO4 required to make the buffer. Also we will be unable to produce a basic pH of 7.1. Hence this cannot the best method.

(B) Here we are taking only K2HPO4 and H3PO4. Since both are acids we are unable to get the desired concentration of KH2PO4 required to make the buffer. Also we will be unable to produce a basic pH of 7.1. Hence this cannot the best method.

(C) This method contains both KH2PO4 andK2HPO4 of desired concentration. However we will be unable to get exact pH of 7.1 if there is any error in the measurement of the required amount of the salt. We can use this method but not the best one.

(D) Here we are taking KH2PO4 and KOH. KOH will react with KH2PO4 to form K2HPO4. Hence we can get the desired concentration of KH2PO4 and K2HPO4. Also we can get an exact basic pH of 7.1 by adjusting the KOH solution. Hence method D should the the best method.


Related Solutions

A liter of 0.050M pH 7.10 potassium phosphate buffer is needed for a certain experiment. The...
A liter of 0.050M pH 7.10 potassium phosphate buffer is needed for a certain experiment. The following four methods were proposed for the preparation of the buffer. Using the pKa value given in your textbook, which of these methods should be used and what is wrong with the other proposals? Letter of method which should be used: __________________________ (2 points) (A) A 0.050M sample of K2HPO4 was dissolved in about 800.0mL of water, and the pH was adjusted to 7.10...
A liter of pH 7.20 phosphate buffer is needed for a certain experiment. The Henderson-Hasselbach equation...
A liter of pH 7.20 phosphate buffer is needed for a certain experiment. The Henderson-Hasselbach equation will be sufficiently accurate for your determination of pH. The pK’s for possibly relevant phosphate species are: H3PO4 ↔ H2PO4- + H+ pK = 2.15 H2PO4- ↔ HPO4-2 + H+ pK = 7.20 HPO4-2 ↔ PO4-3 + H+ pK = 12.4 0.100 Moles of H3PO4 were dissolved in about 800 mL of water, and the pH was adjusted to 7.20 using a standardized pH...
Prepare 100mL of a 0.1M potassium phosphate buffer of pH 6.5. You have K2HPO4 and KH2PO4...
Prepare 100mL of a 0.1M potassium phosphate buffer of pH 6.5. You have K2HPO4 and KH2PO4 in solid/salt form available to mix with water to create this buffer. Final answer must include how many grams of each salt to mix with how many mL of water to prepare the buffer. I started this problem and got as far as using the Henderson-Hasselbalch equation and that the ratio of [HPO42-] to [H2PO4-] is 0.48 moles to 1 mole.
How would you prepare 10L of 0.045M potassium phosphate buffer, pH 7.5? You can use the...
How would you prepare 10L of 0.045M potassium phosphate buffer, pH 7.5? You can use the Henderson-Hasselbalch equation to calculate how much of each chemical species. This ratio will tell you how much of the 10L comes from the [A-} species, and how much from the [HA] species, giving you the volumes for your next calculation. Then you have to figure out how much of each to weigh out to make that calculated volume at 0.045M for each species. The...
A common buffer used in biochemistry is a pH 7.4 phosphate buffer. Such a buffer costs...
A common buffer used in biochemistry is a pH 7.4 phosphate buffer. Such a buffer costs $53.00 for 100.00 mL from a common chemical supply house. Therefore, it is useful to know how to prepare such a buffer yourself! Imagine that you have available 0.1 M KH2PO4 and solid K2HPO4. Describe in detail how you would prepare 100.00 mL of a pH 7.4 phosphate buffer using these two reagents. **Hint: You need to identify what is your acid and what...
What volume of 60.0 mM potassium maleate buffer, pH 6.5, would be needed to make 0.250...
What volume of 60.0 mM potassium maleate buffer, pH 6.5, would be needed to make 0.250 liters of 10.0 mM maleate buffer containing 1.50 M ammonium nitrate? (6 pts.) 41.7 mL What mass of ammonium nitrate (F.W. 80.04) would be needed? 30.0 g In the above example, how many -fold has the original buffer been diluted?
Prepare a potassium phosphate stock solution with a pka value of 7.2 and ph value of...
Prepare a potassium phosphate stock solution with a pka value of 7.2 and ph value of 7.0. calculate the amount of monobasicpotassium phosphate amount required for 100ml of a 1M solution and th amount of dibasic phosphate amount required for 25ml of a 1M solution?
If pH of 1 liter of a 1.0mM carbonate buffer is 7.0, what is actual number...
If pH of 1 liter of a 1.0mM carbonate buffer is 7.0, what is actual number of moles of H2CO3 and HCO3? (pk=6.37)
One liter of a 0.1M Tris buffer (pKa=8.3) is adjusted to a pH of 2.0. A)...
One liter of a 0.1M Tris buffer (pKa=8.3) is adjusted to a pH of 2.0. A) What are the concentrations of the conjugate base and weak acid at this pH? B) What is the pH when 1.5mL of 3.0M HCl is added to this buffer? Is Tris a good buffer at this pH? Why? C) What is the pH when 1.5mL of 3.0M NaOH is added to this buffer?
A) Consider how best to prepare one liter of a buffer solution with pH = 4.94...
A) Consider how best to prepare one liter of a buffer solution with pH = 4.94 using one of the weak acid/conjugate base systems shown here. Weak Acid Conjugate Base Ka pKa HC2O4- C2O42- 6.4 x 10-5 4.19 H2PO4- HPO42- 6.2 x 10-8 7.21 HCO3- CO32- 4.8 x 10-11 10.32 How many grams of the potassium salt of the weak acid must be combined with how many grams of the potassium salt of its conjugate base, to produce 1.00 L...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT