Calculate the pH of a buffer solution obtained by dissolving 13.0 g of KH 2 PO 4 ( s ) and 32.0 g of Na 2 HPO 4 ( s ) in water and then diluting to 1.00 L

7.58

Explanation:

Given that

A buffer solution obtained by dissolving 13.0 g of KH₂PO₄ and 26.0 g of Na₂HPO₄ in water and then diluting to 1.00 L.

pKa = 7.21.

To get the PH we are going to use Henderson - Hasselblach equation:

PH = Pka + ㏒ [A/AH]

when the molar mass of Na2HPO4 = 142 g/mol 

and A is the conjugate base HPO4-- ions so,

∴[A] = 32g / 142 g/mol 

       = 0.225 M

and when the molar mass of KH2PO4 = 136 g/mol

and AH is the weak acid H2PO4- ions so,

∴[AH] = 13 g / 136 g/mol 

         = 0.096 M

and when we have the Pka value of H3PO4 = 7.21 

so, by substitution:

∴ PH = 7.21 + ㏒ (0.225 / 0.096)

        = 7.58

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

thats a little if u want the answer just look it up on google thats what i do haha : )