A number of phosphate buffers will be prepared during this lab period using the

Question

A number of phosphate buffers will be prepared during this lab period using the concepts and calculations presented in CHEM 4610. For H_3 PO_4 use pK1 = 2.3; pK2 = 6.9; and pK3 = 12.3. Always use distilled water to prepare solutions. Prepare 100 ml of 0.05 M phosphate buffer, pH 6.3, using solid NaH_2 PO_4 and solid Na_2 HPO_4. (See the containers for actual formulas and molecular weights.) Calculate the amounts of the two solids needed for this buffer. Weigh the solids on the analytical balance and transfer to a 100 ml volumetric flask. Fill the flask about 2/3 full with water, dissolve all the solid then fill to the mark and mix well. Measure and record the pH of this buffer with MEASURENET. Prepare 100 ml of 0.05 M phosphate buffer, pH 6.3, using 0.05 M NaH_2 PO_4 and 0.05 M Na_2 HPO_4. Calculate the volume of each of the two solutions needed for this buffer. Combine these volumes in a beaker using a 100 ml graduated cylinder as a measuring device. Mix well. Measure and record the pH of this buffer with MEASURENET. Prepare 250 ml of 0.02 M phosphate buffer in which [H_2 PO_4]^- = [HPO_4]^=. Calculate the volumes of 2 M phosphoric acid and 1 M NaOH required. To a 250 ml volumetric flask containing about 100 ml of water, transfer the required volumes of 2 M phosphoric acid and 1 M NaOH, then fill to the mark with water and mix well. Measure and record the pH of the buffer with MEASURENET. I. Prepare 250 ml of 0.05 M phosphate buffer, pH 7.4, using 2 M H_3 PO_4 and 1 N NaOH. Use a 5 ml Mohr pipet to measure the volume of 2 M H_3 PO_4 needed. Use a 50 ml buret to measure the volume of 1 N NaOH needed. To a 250 ml volumetric flask containing about 100 ml of water, transfer the required volumes of 2 M phosphoric acid and 1 M NaOH, then fill to the mark with water and mix well. Measure and record the pH of this buffer with MEASURENET. Keep this buffer for (II), below. II. Transfer 100.00 ml of the buffer prepared in (I) above into a 150 ml beaker using a buret. First, clean the buret. This is not as easy as it sounds because NaOH adheres to glass fairly well. So, proceed by rinsing the buret with water followed by a rinse with a little dilute HC1 (run the HC1 through the buret tip during this process). This neutralizes any residual NaOH and the resulting salts and HC1 can then easily be rinsed out with water. After this has been done, rinse the buret with some of the buffer from (I), drain the buret well then fill it with the buffer and you are ready to dispense the buffer. To this 100.00 ml of buffer, add 1000 mu I of 2.00 N HCl. Mix well. Assume no volume change and calculate the new pH of the buffer. Measure and record the pH of this resulting buffer solution with MEASURENET.

Explanation / Answer

PART 1.

V = 100 mL of M = 0.05 M of Phosphate Buffer, pH = 6.3 NaH2PO4 and Na2HPO4

Note that in solution

NaH2PO4 --> Na+(aq) + H2PO4-(aq)

and

H2PO4-(aq) <--> H+(aq) + HPO4-2(aq)

also the salt Na2HPO4:

Na2HPO4(aq) --> 2Na+(aq) + HPO4-(aq)

therefore, there is a common ion effect, the HPO4-; this will be a buffer

acid = H2PO4-(aq)

conjugate base = HPO4-2(aq)

the pH of a buffer, by henderson hasselbach equation

pH = pKa2 + log([conjguate base]/[acid]

note that this is NOT H3PO4, but the second ionization, that is H2PO4-, so choose pKa2

pKa2 = 7.21

and if pH = 6.3

substitute data

6.3 = 7.21 + log([conjguate base]/[acid])

[conjguate base]/[acid] = 10^(6.3-7.21)

[conjguate base]/[acid] = 0.1230

[conjguate base] = 0.1230*[acid]

note that we know that the total concentration (0.05 M), that is

[conjguate base] + [acid] = 0.05 M

so we have 2 equations, 2 unkown, solve them

substitute [conjguate base] = 0.1230*[acid] in equation 2

0.1230*[acid] + [acid] = 0.05 M

[acid] = (0.05)/(1.123) = 0.0445 M

[conjguate base] = 0.1230*[acid] = 0.1230*0.0445 = 0.0054735 M

but we need mass, so let us get moles from concentration

mol = M*V

V = 100 mL or 0.1 L

mol of conjugate = M1*V1 = 0.0054735*0.1 = 0.00054735 mol

mol of acid = M2V2 = 0.0445*0.1 = 0.00445 mol

calculate mass, don't forget to relate mol to its salt

mol of conjugate = 0.00054735 mol = mol of Na2HPO4

mass = mol*MW of Na2HPO4

mass of Na2HPO4 = 0.00054735 mol *141.96 g/mol = 0.077 g of Na2HPO4

mol of acid = 0.00445 mol = mol of NaH2PO4

mass of NaH2PO4 = 0.00445 mol *119.98 g/mol = 0.533911 g of Na2HPO4

answer to "mass of species"

0.077 g of Na2HPO4 and 0.533911 g of Na2HPO4

Note that after dilution, the pH should remain the same, since we are having the same ratio of moles, that is

pH = pKa + log([conjugate]/[acid])

since [Concentration] = mol/Vmix

Vmix is the same for conjugate and acid, so

log([conjugate]/[acid]) =  log((mol of conjugate/Vmix)/(mol of acid / Vmix))

the Vmix cancel each other so

log(mol of conjugate/mol of acid) is what really matters

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