Look at HA ionization.
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Solve for (H^+) = Ka(HA)/(A^-)
Note that 700 mL is exactly half way to the equivalence point; therefore, we know (HA) untitrated = (A^-) from the half titrated. Thus pKa = pH.
At th stoichiometric point you know mols HCl = M x L = 0.1M x 1.4L = 0.14mols.
That represents 15.00 g;
grams = moles/molar mass. Solve for molar mass.
15.00 g/molar mass = mols NaA and
M = moles/L at this point. I see 1.4L + 100 mL initially used.M = ?
............A^- + HOH ==> HA + OH^-
initial....?M ..............0....0
change......-x.............x......x
equil....?N-x...............x......x
Kb for A^- = (Kw/Ka for the acid) = (HA)(OH^-)/(A^-)
Substitute from the ICE chart above and solve for x = OH^-, then convert to pH.
A 15.00 g sample of an ionic compound NaA, where A- is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After 700.0 mL of HCl was added, the pH was 5.00. The experimenter found that 1.40 L of 0.100 M HCl was required to reach the stoichiometric point of the titration
what is the molar mass of NaA? and
Calculate the pH of the solution at the stoichiometric point of the titration.
1 answer
1 answer