Phosphoric acid is a triprotic acid with the pkapka values pka1=2.148pka1=2.148 , pka2=7.198 pka2=7.198 , and pka3=12.375 pka3=12.375 . you wish to prepare 1.000 l of a 0.06000.0600 m phosphate buffer at ph 7.6407.640 . to do this, you choose to mix the two salt forms involved in the second ionization, nah2po4nah2po4 and na2hpo4na2hpo4 , in a 1.000 l volumetric flask and add water to the mark. what mass of each salt will you add to the mixture?

 Mass of  monosodium phosphate NaH2PO4 =2,40 g and  Mass of  disodium phosphate Na2HPO4= 5, 68 g  

Explanation:

We identify which one is the Acid and which one is the conjugate base and its molecular weight

In this case: the monosodium phosphate  is the Acid [AH] with molecular weight ( 119,98g/mol)  and  

the disodium phosphate is the conjugate base [ A-] with molecular weight ( 141,96 g/mol)

Review the parameters giving:

+Volume of buffer to prepare : 1L

+pH of buffer to prepare: 7,640

+ Molar Concentration of buffer to prepare: [0,06M] That means  [AH] + [ A-] = [0,06M] Equation 1

We use  the Henderson-Hasselbach equation to calculate how much acid and base we will need .

pH = pKa + log ([Base]/[Acid]) = pKa + log ([A-]/[AH])  Equation 2

We choose the pKa  value that is closest to our pH parameter giving. In our casem we choose pKa 7,198, because is closer to our pH 7,640, and we put them in our Equation 2

pH = pKa + log ([A-]/[AH])

7,640  = 7,198 + log ([A-]/[AH])

0,44= log ([A-]/[AH])

2,77 = [A-]/[AH]

2,77 [AH] = [A-]  Equation 3

Now we use Equation 3 in our Equation 1 and make the Sum of equal terms, finally we have the concentration of the Acid

[AH] + [ A-] = [0,06M] Equation 1

2,77 [AH] = [A-]  Equation 3

[AH] + 2,77 [AH]= [0,06M]

3,77 [AH] =[0,06M]

[AH] = [0,02M]

And the Concentration of the conjugate base [ A-], can be calculated from Equation 1 :

[AH] + [ A-] = [0,06M]  

[0,02M] + [ A-] = [0,06M]  

[ A-]  =  [0,04M]  

We calculate the mass of each compound by  multiplying its molecular weight with its molar concentration

mass AH = 119,98g/mol X  0,02mol / L = 2,40 g / L

mass A- = 141,96 g/mol X  0,04mol / L =  5, 68 g / L

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

hope this and sorry i could only come up with one answer!