(d) (4 pts) Using just oxalic acid, which is a solid, and stock solutions of eit

Question

(d) (4 pts) Using just oxalic acid, which is a solid, and stock solutions of either 3.00 M HCla) or 3.00 M (e) (4 pts) If sodium oxalate is used to prepare a buffer solution with pH 4.0, and the concentration of the oxalate (0) (4 pts) Using just sodium oxalate and stock solutions of either 3.00 M HCl(aq) or 3.00 M NaOH(a), how many NaOH(a) how many grams of oxalic acid and milliliters of stock solution are needed to prepare l 00 L of a buffer that has pH 1.5 and contains 0.22 M oxalic acid? ion is 0.16 M, what are the concentrations of the two other species derived from the oxalate ion in this buffer solution? grams of sodium oxalate and milliliters of stock solution are needed to prepare 1.00 L of a buffer that has pH 4.0 and contains 0.16 M oxalate ion?

Explanation / Answer

d) pH = pka1 + log(salt/oxalicacid

   pka1 of oxalicacid = 1.25

no of mol of NaOH must add = x

no of mol of oxalicacid present = 0.22*1 = 0.22 mol

1.5 = 1.25 + log(x/0.22)

x = 0.3912

volume of NaOH must be added = n/M = 0.3912/3 = 0.1304 L

    = 130.4 ml

amount of oxalicacid must take = (0.22+0.3912) *90 = 55 g

e) pH = pka1 + log(salt/oxalicacid)

      pka2 of oxalicacid = 4.27

   4.0 = 4.27+log(0.16/x)

x = 0.298

concentration of oxalicacid formed = x = 0.298 M

for this HCl is added,

so that, concentration of Cl^- = 2x = 2*0.298 = 0.596 M

f) pH = pka1 + log(salt/oxalicacid)

      pka2 of oxalicacid = 4.27

   4.0 = 4.27+log(0.16/x)

x = 0.298

no of HCl must be added = 2x = 2*0.298 = 0.596 mol

volume of HCl must be added = n/M = 0.596/3 = 0.199 L

     = 199.0 ml

no of mol of sodium oxalate required = 0.16+x

                                      = 0.16+0.298

               =   0.458 mol

Amount of sodium oxalate required = 0.458*134 = 61.372 g

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