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ID: 948460 • Letter: H
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home / study / science / chemistry / questions and answers / 1. calculate the amount of na2hpo47h2o (in grams) ... Question 1. Calculate the amount of Na2HPO4 x 7H2O (in grams) needed to make 50 mL of a 0.40 M solution. 2. Calculate the amounts of Na2HPO2 and NaH2PO4 (in grams) needed to prepare 200 mL of a buffer with pH = 8.25 so that the sum of concentrations of (HPO4)2- and (H2PO4)- ions is 0.5 M. pKa1=2.12, pKa2=7.21, pKa2=7.21, pKa3=12.38 3. Write the equations for the ionization of 3-aminophenol. Would this substance be more soluble in water at pH 2.0 or pH 8.0? Would you expect this substance to be absorbed into the bloodstream in the stomach or the intestines? If you could show your work, that would be most helpful. I am lost at what to do.
Explanation / Answer
Let HA= NaH2PO4^- and A^-= NaHPO4^-2
pH=8.25
pKa=7.21. This is the pKa value that you need; the other pKa values are irrelevant.
Henderson-hasselbach equation:
pH=pka+log[A^-]/[HA]
Solving for A^-/HA
10^(pH-pKA)=A^-/HA
10^(8.25-7.21)=A^-/HA
11.0=A^-/HA
Moles of Buffer=molarity*volume
Moles of Buffer=(0.5M)*(0.200L)=0.1 moles of buffer
We have two equations:
1.) A^- + HA=0.100
2.) A^-/HA=11
We have two equations and two unknowns, solve for one unknown using algebra:
Rearrange equation 2 and plug it into equation 1. Solve for HA
11.0HA=A^-
HA +11HA=0.100
12HA=0.100
12HA/12=0.100/12
HA=8.33 x 10^-3 moles
Solving for A^-
0.100-8.33 x 10^-3=9.17 x 10^-2 moles of A^-
Convert moles to mass in g:
9.17 x 10^-2 moles of Na2HPO4 *(141.96g/mol)=13.0g of Na2HPO4
8.33 x 10^-3 moles of NaH2PO4*(119.98g/mol)=0.999g of NaH2PO4
Add the following masses of each to a 200mL volumetric flask and fill with water until the bottom of the meniscus touches the line of the 200mL volumetric flask; shake vigorously.
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