In this experiment I am assigned a pH and must prepare a buffer solution using N
ID: 542999 • Letter: I
Question
In this experiment I am assigned a pH and must prepare a buffer solution using NH3 and NH4CI having that pH. My assigned pH is below: pH-pka +/-1 = 9.25 1 2. Ammonium chloride (NH4Cl, Ka for NH4+-5.6 x 10-10) and ammonia (NH3). The buffer will be prepared by choosing the appropriate acid-base pair, calculating the molar ratio of acid to base that will produce the assigned pH, and then mixing the calculated amounts of the two compounds with enough deionized water to make 200. mL of buffer solution. I NEED HELP ANSWERING THE FOLLOWING QUESTIONS; 3. Determine the ratio of [A VHA] needed in the buffer 4. Which component will be more concentrated, A or HA? Choose a value for the concentration of that component anywhere in the range 0.2-0.6 M. Calculate what the concentration of the other component should be. 5. Calculate the volume (in mL) of 17.5 M CH3CO2H or 14.8 M NH3 needed to make 200 mL of your buffer. Calculate the mass (in g) of solid NaCH3CO2 3H20 or NH4Cl needed to make 200 mL of your buffer.Explanation / Answer
You are preparing a NH4+/NH3 buffer. The acid dissociation reaction is given as
NH4+ (aq) --------> H+ (aq) + NH3 (aq); pKa = 9.25
2) You can prepare a buffer having pH (pKa), i.e, you can prepare buffer having pH in the range 8.25-10.25. Suppose we pick a pH value 9.75.
3) Use the Henderson-Hasslebach equation to find out the ratio [A-]/[HA] where A- = NH3 and HA = NH4+. Plug in the values for pH and pKa and determine [A-]/[HA].
pH = pKa + log [A-]/[HA]
====> 9.75 = 9.25 + log [A-]/[HA]
====> 0.50 = log [A-]/[HA]
====> [A-]/[HA] = antilog(0.50) = 3.162 3.16:1
Therefore, the ratio is [A]:[HA] = 3.16:1 (ans).
4) We have determined the ratio as above. Therefore,
[A-] = 3.16*[HA]
Consequently, A-, i.e, NH3 is the more concentrated component of the buffer (ans).
As per the problem, let us assume [NH3] = 0.5 M. Therefore,
[NH3] = 3.16*[NH4+]
====> 0.5 M = 3.16*[NH4+]
====> [NH4+] = (0.5 M)/(3.16) = 0.1582 M 0.16 M (ans).
5) We wish to prepare 200 mL buffer with NH3 concentration 0.5 M. Use the dilution equation to find out the volume of stock NH3 solution required.
M1*V1 = M2*V2 where M1 = 14.8 M (given), V2 = 200 mL and M2 = 0.5 M.
===> (14.8 M)*V1 = (0.5 M)*(200 mL)
===> V1 = 6.7567 mL 6.76 mL.
Take 6.76 mL of 14.8 M NH3 solution and dilute to 200.0 mL mark with deionized water.
Molar mass of NH4Cl = (1*14.0067 + 4*1.008 + 1*35.453) g/mol = 53.4917 g/mol.
Moles of NH4Cl in 200 mL buffer = (200 mL)*(1 L/1000 mL)*(0.16 M)*(1 mol/L/1 M) = 0.032 mole.
Mass of NH4Cl required = (0.032 mole)*(53.4917 g/mol) = 1.7117 g (ans).
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