10. The other day, I reached out to my sensitive and artistic side. I began to c
ID: 104865 • Letter: 1
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10. The other day, I reached out to my sensitive and artistic side. I began to compose biochemistry haikus. Here is one about pH: resist pH change acids and their conjugates equilibriunm You need to make 1 liter of a 100mM phosphate buffer at a pH of 6.8. The pKa for HyPO,is 2.1. The pka for NaH PO, is 7.2. The pKa for Na,HPOs is 12.3. NaH2PO, is referred to as sodium phosphate monobasic while Na,HPO, is referred to as sodium phosphate dibasic. You have a stock solution of the monobasic at 2M, and a stock solution of the dibasic at 1M, and a stock solution of phosphoric acid at 0.5M. To make 1L this buffer, how much volume of A) water, B) stock dibasic, C) stock monobasic, and D) stock phosphoric acid will you need to mix? (Notice, on the final, all of the highlighted values could change; zero might be an answer for some of them)Explanation / Answer
Consider the step-wise dissociation of phosphoric acid, H3PO4 as below.
H3PO4 (aq) <=====> H+ (aq) + H2PO4- (aq); pKa1 = 2.1
H2PO4- (aq) <======> H+ (aq) + HPO42- (aq); pKa2 = 7.2
HPO42- (aq) <======> H+ (aq) + PO43- (aq); pKa3 = 12.3
We wish to prepare a buffer that has a pH of 6.8. The desired pH is close to the pKa2 of phosphoric acid and hence, we will choose the buffer system as H2PO4-/HPO42-. Since these anions do not exist in nature, we will choose salt solutions of the anions, sodium phosphate monobasic, NaH2PO4 and sodium phosphate dibasic, Na2HPO4.
Use the Henderson-Hasslebach equation to find out the ratio of the monobasic and the dibasic in the buffer. The Henderson-Hasslebach equation is written as
pH = pKa2 + log [Na2HPO4]/[NaH2PO4] (square braces denote molar concentrations)
====> 6.8 = 7.2 + log [Na2HPO4]/[NaH2PO4]
====> -0.4 = log [Na2HPO4]/[NaH2PO4]
====> [Na2HPO4]/[NaH2PO4] = antilog(-0.4) = 0.398
====> [Na2HPO4] = 0.398*[NaH2PO4] ……(1)
The total phosphate concentration is given as 100 mM, i.e,
[Na2HPO4] + [NaH2PO4] = 100 mM = (100 mM)*(1 M/1000 mM) [1 M = 1000 mM]
====> [Na2HPO4] + [NaH2PO4] = 0.1 M
====> [NaH2PO4] + 0.398*[NaH2PO4] = 0.1 M
====> 1.398*[NaH2PO4] = 0.1 M
====> [NaH2PO4] = (0.1 M)/1.398 = 0.071530 M 0.0715 M.
Therefore, [Na2HPO4] = 0.398*[NaH2PO4] = 0.398*(0.0715 M) = 0.028457 M 0.0285 M.
We have stock solutions of NaH2PO4 and Na2HPO4 available. We can easily prepare the buffer solution by taking aliquots from the stock solutions. We will use the dilution equation to find the volume(s) of the stock solutions required. The dilution equation is given as
M1*V1 = M2*V2 where M1 = concentration of the stock solution; M2 = concentration of the species in the buffer solution, V1 = volume of stock solution taken and V2 = volume of the buffer solution = 1 L = (1 L)*(1000 mL/1 L) = 1000 mL.
NaH2PO4: (2 M)*V1 = (0.0715 M)*(1000 mL)
====> V1 = (0.0715 M)*(1000 mL)/(2 M) = 35.75 mL
Na2HPO4: (1 M)*V1 = (0.0285 M)*(1000 mL)
====> V2 = (0.0285 M)*(1000 mL)/(1 M) = 28.50 mL
The buffer is prepared by adding (A) [1000 – (35.75 + 28.50)] mL = 935.75 mL water, (B) = 28.50 mL of 1 M stock dibasic and C) 35.75 mL of 2 M stock monobasic. (D) 0.00 mL of stock phosphoric acid is required to prepare the buffer.
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