When 50.0 mL of 0.50 M formic acid is titrated with 0.50 M sodium hydroxide, cal

Question

When 50.0 mL of 0.50 M formic acid is titrated with 0.50 M sodium hydroxide, calculate the pH for each of the following:

Be sure to account for volume changes at each step.

a) Before any of the titrant (NaOH solution) is added

b) After 15.0 mL of the titrant has been added

c) When half of the formic acid has been neutralized

d) At the equivalence point

e) When 75.0 mL NaOH has been added

f) Select a good indicator for this titration (see page 746 in your textbook).

g) Describe how to prepare a buffer solution from HCO2H and NaHCO2 having a pH of 4.60.

Explanation / Answer

a) [H+] = sqrt (Ka C) = sqrt (1.8 * 10 ^ -4 * 0.50 ) = 9.48 * 10 ^ -3 M
pH = 2.022

b) Moles of acid = 0.025
Moles of base = 0.0075
So excess moles of acid = 0.0175
[H+ ] = 2.69 * 10 ^ -7
pH = 6.569

c) Moles of acid remaining = 0.0125
[H+] = 1.66 * 10 ^ -7

pH = 6.778

d) pH > 7

e) Moles of acid = 0.025
Moles of base = 0.0375
Moles of base in excess=0.0125
[OH-] = 10 ^ -7
pH = 7

f) Bromthymol blue is the best indicator

g)pH = pKa + log (salt/acid)
4.60 = 4.74 + log (salt/acid)
log(salt/acid) = -0.144
salt/acid = 0.71
If u take NaHCO2 and HCOOH in the ratio of 0.71 then a abuffer of pH 4.60 can be obtained.

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