Chemistry help....pH of a salt solution?

Question

Trisodium phosphate (Na3PO4) is available in hardware stores as TSP and is used as a cleaning agent. The label on a box of TSP warns that the substance is very basic (caustic or alkaline). What is the pH of a solution containing 99.9 g of TSP in a liter of solution?


Can someone walk me through this problem?

Answer 1

Phosphoric acid is a wead triprotic acid with pKa1=2.12, pKa2= 7.21 and pKa3=12.67
Thus the anion PO4(-3) is a weak base that hydrolyzes according to the reactions

PO4(-3) + H2O <=> HPO4(-2) +OH(-) with Kb1=Kw/Ka3
HPO4(-2) + H2O <=> H2PO4(-) + OH(-) with Kb2=Kw/Ka2
H2PO4(-) + H2O <=> H3PO4 +OH(-) with Kb3=Kw/Ka1

For simplicity, we will assume that only the first reaction occurs. This is justified since the Kb2 and Kb3 are 5 and 10 orders of magnitude (respectively) smaller (this you see by looking the pKa: pKa=-logKa, so a difference of 1 pKa unit corresponds to one order of magnitude; also pKb=pKw-pKa, that's why the bigger pKa value the smaller the pKb)

The molecular weight of TSP is 164 g/mole, so the initial concentration is C= 99.9 (g/L) /164 (g/mole) =0.609 mole/L
Na3PO4 dissociates completely to Na+ and PO4(-3), thus the initial concentration of PO4(-3) is also 0.609 mole/L
Now we set up an ICE table

.. .. .. .. .. .. PO4(-3) + H2O <=> HPO4(-2) +OH(-)
Intial .. .. .. .. 0.609
React .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x
At equil .. .. 0.609-x .. .. .. .. .. .. .. .. x .. .. .. .. .. x

Kb1= [HPO4(-2)][OH-] / [PO4(-3)]= Kw/Ka3 =>
Kw/Ka3= x^2/(0.609-x)

Let's do the assumption that 0.609>>x so that 0.609-x=0.609.
Then the equation is simplified to
Kw/Ka3=x^2/0.609 =>
x= squareroot( 0.609*Kw/Ka3)
but Ka3=10^-pKa3 so we substitute
x = SQRT( 0.609*(10^-14) / (10^-12.67) ) =0.169 which is not much smaller than 0.609, so our assumption is not valid and we need to solve the quadratic equation

x^2/(0.609-x)= 10^-14 /10^-12.67 =10^-1.33= 0.04677 =>
x^2+ 0.4677x -0.28483 =0
x1= 0.3488
x2= -0.81653<0 rejected

so pH= 14-pOH=14 -(-logx) =14-(-log(0.3488)) =13.54

Answer 2

Ka of H3PO4 is required.