H3PO4 ==> H^+ + H2PO4^-
K1 = (H^+)(H2PO4^-)/(H3PO4)
You are given H3PO4 and pH. For all practical purposes, although H3PO4 has k1, k2, and k3, the first ionization of k1 is so large in comparison with the others that you may consider this a monoprotic acid, at least for the purposes of this problem. Use pH to find (H^+), set up and ICE chart, substitute into Ka expression and solve for Ka.
A 8.6 X10^-3M Solution of H3PO4 has a Ph=2.30. what is Ka for H3PO4?
Have no clue how to solve this problem
3 answers
what is the molarity of an aqueous solution containing 40 g of glucose (c6h12o6) in 1.5 L in solution
molarity = moles/L soln.
moles = 40/molar mass.
L soln = 1.5 L
Plug and chug.
moles = 40/molar mass.
L soln = 1.5 L
Plug and chug.
2 answers