it\'s the post lab question. I\'m confused how to do this. I solved for the mola
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it's the post lab question. I'm confused how to do this. I solved for the molar ratio of B to A which is 2.01:1 and I don't know where to go from there
1. The goal of the buffers part of the experiment 2. Methodology used in the buffers part of the experiment lationship between different ratios of acid/conjugate base on pH w/ ra 4. Effect of adding strong acid to water w/ rationale 5. Effect of adding strong acid to buffer w/rationale 6. Effect of diluting acid on pH w/rationale 7. Effect of diluting base on pH w/rationale 8. Effect of diluting buffer on pH w/rationale 9. How understanding the relationship between buffer components and pH is helpful POST-LAB QUESTION (7 points) this lab, you prepared a buffer and then determined its pH, but in a laboratory setting it is usually the other way around- you need a buffer at a particular pH, and this dictates its preparation. Describe how you would prepare 100. mL of 0.1000 phosphate buffer at pH 6.5, given appropriate glassware, sodium phosphate, monobasic (NaH2PO, pKa = 7.20) and sodium phosphate, dibasic (Nall Po, pKa = 12.35) at 25°C Your answer should include calculations. (Note: The moles of each component must add to the buffer moles. Le. A+ HA 0.250 mol) InExplanation / Answer
Consider the ionization of phosphoric acid as below.
H3PO4 (aq) --------> H+ (aq) + H2PO4- (aq); pKa1 = 2.12
H2PO4- (aq) --------> H+ (aq) + HPO42- (aq); pKa2 = 7.20
HPO42- (aq) ---------> H+ (aq) + PO43- (aq); pKa3 = 12.35
The pH of the solution is 6.5; hence, we shall choose NaH2PO4 and Na2HPO4. Determine the ratio of NaH2PO4 and Na2HPO4 in the buffer using the Henderson-Haslebach equation as
pH = pKa + log [Na2HPO4]/[NaH2PO4]
====> 6.5 = 7.20 + log [Na2HPO4]/[NaH2PO4]
====> log [Na2HPO4]/[NaH2PO4] = 6.5 – 7.20 = -0.7
====> [Na2HPO4]/[NaH2PO4] = antilog (-0.7) = 0.1995 0.20
====> [Na2HPO4] = 0.2*[NaH2PO4] ……(1)
The total phosphate concentration is 0.1000 M; therefore,
[Na2HPO4] + [NaH2PO4] = 0.1000 M
=====> 0.2*[NaH2PO4] + [NaH2PO4] = 0.1000 M
=====> 1.2*[NaH2PO4] = 0.1000 M
=====> [NaH2PO4] = (0.1000 M)/(1.2) = 0.0833 M
Therefore, [Na2HPO4] = 0.2*[NaH2PO4] = 0.2*(0.0833 M) = 0.0167 M.
We have 100 mL of the buffer; hence, determine the mole(s) of the components present in the buffer.
Mole(s) of NaH2PO4 = (100 mL)*(1 L/1000 mL)*(0.0833 M) = 0.00833 mole.
Mole(s) of Na2HPO4 = (100 mL)*(1 L/1000 mL)*(0.0167 M) = 0.00167 mole.
The molar masses are required.
NaH2PO4: 119.98 g/mol
Na2HPO4: 141.96 g/mol
Mass of NaH2PO4 required = (0.00833 mole)*(119.98 g/mol) = 0.9994 g.
Mass of Na2HPO4 required = (0.00167 mole)*(141.96 g/mol) = 0.2371 g.
In order to prepare the buffer, weigh out 0.9994 g NaH2PO4 and 0.2371 g Na2HPO4 in a 100 mL volumetric flask and make up to the mark with deionized water.
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