Given 1L of 0.1M stock solutions of NaH2PO4 and Na2HPO4, and ample water, describe in as...

Given 1L of 0.1M stock solutions of NaH2PO4 and Na2HPO4, and ample water, describe in as much detail as possible how you would go about making a 0.05M solution of phosphate buffer with a pH=6.75?

Solutions

Expert Solution

Given 1L of 0.1M stock solutions of NaH2PO4 and Na2HPO4, and ample water, describe in as much detail as possible how you would go about making a 0.05M solution of phosphate buffer with a pH=6.75

Here in this question Na2HPO4 will behave as base and NaH2PO4 will behave as acid.

Na2HPO4àà 2Na⁺ + HPO4^2-

NaH2PO4àà Na⁺ + H2PO4^-

And then; according to Bronsted-Lowry's theory:

H2PO4^- àà H⁺ + HPO4^2-

Complete equation can be written as Na2HPO4 àà Na⁺ + H⁺ + HPO4^2-

We can use handerson hasselbalch equation,

pH = pKa +log [Na2HPO4]/[NaH2PO4]

pKa of phosphate buffer is 7.21

6.75 = 7.21 + log [Na2HPO4]/[NaH2PO4]

log [Na2HPO4]/[NaH2PO4] = -0.46

[Na2HPO4]/[NaH2PO4] = -antilog 0.46

[Na2HPO4]/[NaH2PO4] =0.3467

We need [Na2HPO4] + [NaH2PO4] = 0.05

By solving above two equations we get,

[Na2HPO4]= 0.3467[NaH2PO4]

0.3467[NaH2PO4] + [NaH2PO4] = 0.05

[NaH2PO4] = 0.05/1.3464 = 0.037

[Na2HPO4] = 0.05-0.037 = 0.012

This means you need 0.037 moles of NaH2PO4 and 0.012moles of Na2HPO4 in 1lit of solution

Now moles= molarity *volume

Therefore volume= 0.037/0.1=0.37 liter of NaH2PO4

And = 0.012/0.1= 0.12liter of Na2HPO4

370ml of NaH2PO4 and 120liter of Na2HPO4 should be mix in 1 liter of water to get desired concentration of buffer with pH 6.75

But molarity of the solution is given as 0.1M

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

using NaH2PO4 and Na2HPO4 both are stock solutions of .2 M 3) Calculate volumes of both...

using NaH2PO4 and Na2HPO4 both are stock solutions of .2 M 3) Calculate volumes of both solutions required to prepare 150 mL 0.1 M phosphate buffer given pKa 6.8 and we want the pH to be 7.0 im not sure if you need the gram values from the preceding problems for 50 ml solutions.. we went over this today the formula was something like (x/.1-x)=10^7-6.8 then somehow the value of mLs was calculated and a part of the solution was...

You have 1 M solutions of H3PO4, NaH2PO4, Na2HPO4, Na3PO4 (no NaOH). => Using only two...

You have 1 M solutions of H3PO4, NaH2PO4, Na2HPO4, Na3PO4 (no NaOH). => Using only two of these solutions, how would you prepare 150 mL of 1 M phosphate buffer at pH 3.0? A) 131.4 mL of NaH2PO4 + 18.6 mL H3PO4 B) 50 mL Na2HPO4 + 100 mL Na3PO4 C) 75 mL Na2HPO4 + 75 mL Na3PO4 D) 18.6 mL of NaH2PO4 + 131.4 mL H3PO4

Calculate the amount of stock reagent(s) and water required to make the following solutions. A)2.5 L...

Calculate the amount of stock reagent(s) and water required to make the following solutions. A)2.5 L of 25 mM NaCl (from 250 mM stock), 1 mM EDTA (from 0.5 M stock) B)200 mL of 0.25 M Sodium Citrate (from dry powder), 20 mM SDS (from 0.5 M stock), 10% glycerol C)3.0 L of 25% Saroksyl

Using the stock solution provided (7% NaCl) and water, prepare the following solutions: i) 1 mL...

Using the stock solution provided (7% NaCl) and water, prepare the following solutions: i) 1 mL of NaCl solution with an osmolarity of 50 mOsM ii) 1 mL of NaCl solution with an osmolarity of 300 mOsM iii) 1 mL of NaCl solution with an osmolarity of 1200 mOsM

3) Starting with two stock buffer solutions (A and B) listed below, describe how you would...

3) Starting with two stock buffer solutions (A and B) listed below, describe how you would mix them together to obtain the solutions i and ii. Write your answers to 3 significant figures. i) 400 mL of 0.35 M NaCl, 10 mM Tris buffer (pH 8.0) ii) 600 mL of 0.9 M NaCl, 10 mM Tris buffer (pH 8.0) Stock buffer A: 3 M NaCl, 10 mM Tris buffer (pH 8.0) Stock buffer B: 0.1 M NaCl, 10 mM Tris...

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

. Given 100mg/L benzoic acid and 200 mg/L caffeine stock solutions, explain how you will make...

. Given 100mg/L benzoic acid and 200 mg/L caffeine stock solutions, explain how you will make 2, 4, 6, 8, and 10mg/L benzoic acid solutions and 4, 8, 12, 16, and 20 mg/L caffeine solutions. Assume that you will have 50 and 100 mL volumetric flasks and your usual variety of pipet sizes and list the exact volume glassware you would use for each of these dilutions. Be sure to finish this problem and list exact glassware, the answer here...

. Given 100mg/L benzoic acid and 200 mg/L caffeine stock solutions, explain how you will make...

Once the stock solutions are prepared, describe how you would produce your buffers (four). Your explanation...

Once the stock solutions are prepared, describe how you would produce your buffers (four). Your explanation should be detailed and quantitative. For my stock solution: KH2PO4 = (136.08 g/mol)(1M)(100.00ml/1000ml)= 13.608 gm K2HPO4 = (174.17 g/mol)(1M)(100.00ml/1000ml)= 17.4178 gm KH2PO4= (136.08 g/mol)(1M)(250.00ml/1000ml)= 34.02 gm K2HPO4= (174.17 g/mol)(1M)(250.00ml/1000ml)= 43.54 gm KH2PO4= (136.08 g/mol)(1M)(25.00ml/1000ml) = 3.402 gm K2HPO4= (174.17 g/mol)(1M)(25.00ml/1000ml) = 4.35425 gm KH2PO4= (136.08 g/mol)(1M)(10.00ml/1000ml)= 1.3608 gm K2HPO4= (174.17 g/mol)(1M)(10.00ml/1000ml)= 1.74178 gm KH2PO4= (136.08 g/mol)(1M)(5.00ml/1000ml)= 0.6804 gm K2HPO4=...

You are given a stock soltuion of (-)-epicatechin (EC, MW=290.26 g/mol) in water (0.94 mg/mL). You...

You are given a stock soltuion of (-)-epicatechin (EC, MW=290.26 g/mol) in water (0.94 mg/mL). You prepare a series of solutions by serial dilution as follows: (1) diluting 0.5 mL of the stock solution to 10 mL with water (solution A), (2) diluting 1.5mL of solution A with 4 mL of water (solution B), and (3) diluting 3 mL solution B with 9 mL water (Solution C). What is the concentration of solution C (mg/mL and microM)? What is the...

Subjects