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?

Solutions

Expert Solution

To calculate the amount of monobasic potassium phosphate , molecular weight of monobasic potassium phosphate= 136.086g

Required amount 100ml(0.100L) of 1M (whereM=moles/L) solution:

= 0.100L× 1Mol monobasic potassium phophate× 136.086g/L× Mol

= 13.608g

13.608g is required for 100ml of 1M solution to be prepared.

To calculate the amount required for dibasic phosphate, molecular weight of dibasic phosphate (Na2HPO4) = 141.96g

Required amount is 25ml( 0.025 Litres) of 1M solution:

= 0.025L×1Mol dibasic phosphate×141.96g/L×mol

= 3.549g

3.549g of dibasic phosphate is required for 25ml of 1M solution to be prepared.

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

a) Calculate the pH of a 0.045 M potassium chlorite solution. (The pKa for chlorous acid...

a) Calculate the pH of a 0.045 M potassium chlorite solution. (The pKa for chlorous acid is 1.92) b) Calculate the pH of a solution that is 0.35 M hydroxylamine and 0.50 M hydroxylammonium chloride, (The pKb for hydroxylamine is 7.96)

An enzyme-catalyzed reaction was carried out in a solution buffered with 0.03 M phosphate, PH 7.2....

An enzyme-catalyzed reaction was carried out in a solution buffered with 0.03 M phosphate, PH 7.2. As a result of the reaction, 0.004mole/liter of acid was formed a) What is the pH at the end of the reaction.? b) what would the pH if No buffer were present? Answers for questions a and b are pH = 6.9 and pH =2.4 respectively. BUT HOW DID THEY GET THOSE pH?

You prepare a 0.200M solution of a monoprotic buffer at pH 7.22. The pKa of this...

You prepare a 0.200M solution of a monoprotic buffer at pH 7.22. The pKa of this buffer is 7.53 at room temperature. a. What is the molarity of the acid and conjugate base necessary to make the buffer. [HA]=? [A-]=? b. A separate 1.0 M stock solution of the acid and conjugate base is made/ How many mL of each will you use to prepare 1.0 L of the buffer in part a? mL of HA=? mL of A-=? c....

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.

Design a procedure to prepare 75 ml of 0.10 M potassium dihydrogen phosphate solution (KH2PO4) and...

Design a procedure to prepare 75 ml of 0.10 M potassium dihydrogen phosphate solution (KH2PO4) and 75 ml of 0.10 M dipotassium hydrogen phosphate solution (K2HPO4) from the solids. (Hint: What mass of the above solids is required to make 75 ml of a 0.10 M solution?)

The molar solubility of lead phosphate in a 0.180 M potassium phosphate solution is _______ M.

The molar solubility of lead phosphate in a 0.180 M potassium phosphate solution is _______ M. The maximum amount of lead sulfate that will dissolve in a 0.299 M ammonium sulfate solution is _______M. The molar solubility of lead phosphate in a 0.259 M lead nitrate solution is _______ M.

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...

Given stock solutions of 25 mM nitrophenyl phosphate, and 1.5 M tris,pH 7.2, 7.5 mM MgCl2....

Given stock solutions of 25 mM nitrophenyl phosphate, and 1.5 M tris,pH 7.2, 7.5 mM MgCl2. If you mix 0.1ml of the nitrophenyl phosphate, 0.7 ml of the tris solution and 0.2 ml of water to bring the total volume to 1 ml, what is the final concentrations of each of the components in the final solution? Show work Nitrophenyl phosphate in mM Tris in M Magnesium chloride in mM

The pH of Chloroacetic acid solution is 1.71 Based on the pH and the pKa of...

The pH of Chloroacetic acid solution is 1.71 Based on the pH and the pKa of 2.85 what percentage of the Chloroacetic acid is in HA(protonated) form

What is the pH of a 0.15 M solution of potassium nitrite (KNO2)? (The Ka value...

What is the pH of a 0.15 M solution of potassium nitrite (KNO2)? (The Ka value for nitrous acid (HNO2) is 4.0x10^-4).

Subjects