You have 50 x 0.1 = 5 millimoles NaGlu which will require 100 mL of the 0.05M HCl. (HGlU) = 5 mmols/150 mL = 0.0333
.........HGlu ==> H^+ + Glu^-
I......0.0333.....0......0
C........-x........x......x
E......0.0333-x....x......x
Ka = (H^+)(Glu^-)/(HGlu)
Substitute from the ICE chart and solve for H^+, convert to pH.
........Glu^- + HOH ==> HGlu + OH^-
initial.
Glutamate (Glu–) is the conjugate base form of glutamic acid (HGlu). The Ka of glutamic acid is 5.012 x 10^–5. You titrate 50 mL of 0.10 M sodium glutamate solution with 0.05 M HCl solution. Calculate the pH of the solution at the equivalence point.
Ive been trying to figure this out for hours! Please help me set this up!! thank you so much!!
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