45.0 mL of a 0.10-M formic acid (HCHO2, Ka=1.7x10^-4) solution was titrated with

Question

45.0 mL of a 0.10-M formic acid (HCHO2, Ka=1.7x10^-4) solution was titrated with a 0.20-M NaOH solution. What is the equivalent volume? Calculate which is in excess.

6. What is the pH of the acidic solution before titration?

7. What is the pH of the reaction mixture (in Question 6) after 10.0 mL of NaOH was added?

8. What is the pH of the reaction mixture (titration in Question 6) at the equivalence point?

9. What is the pH of the reaction mixture (titration in Question 6) after 25.0 mL of base was added?

Explanation / Answer

Q1.

First, we need to understand the equivalent volume; that is, the volume needed of NaOH (base) in order to fully react with the acid (formica acid). That is, the momment in which moles of acid = moles of base.

We can do this, but we need the reaction

HCHO2(aq) + NaOH(aq) --> H2O(l) + NaCHO2(aq)

Therefore, ratio is 1:1, meaning 1 mol of acid = 1 mol of base

calculate mol of acid present

mol = M*V

M: molarity/concentration

V: Volume of solution

mol = mol/amount of substance

then

mol of acid = M1*V2

mol of acid = 45*0.1= 4.5 mmol or 4.5*10^-3 mol of acid

now,

since ratio is 1:1 with respect to stochiometry; we need the same amount of base

mol of base = mol of acid = 4.5 mmol of base

then

go backwards

mol = M*V

V = mol/M

V = 4.5/0.2 = 22.5 mL

Volume of base required is about 22.5 mL of NaOH.

NOTE that the questions, "hich is in excess" is not applicable/Valid, since we are looking for the equivalence point. The only way that we will be able to know fi there is excess, is if we had an extra comment such as "we used 30 mL of base" or something similar.

Q6.

before any addition of base, we have only acid

note that this is a weak acid, since we have a Ka value relatively small; meaning it favours acid formation and no ionic species.

then

HCHO2 <-> H+ + CHO2-

Ka = [H+][CHO2-]/[HCHO2]

due to stochiometry, [H+] = [CHO2-] = x

[HCHO2] = M-x = 0.1 - x

note that we need to account for the species in equilibrium (that is, the small fraction of H+ and CHO2- formed)

now, substitute

1.7*10^-4 = x*x / (0.1-x)

solve for x (mathematically, using quadratic formula)

x = 0.004038 is the only valid value

since x = [H+]; then [H+] = 0.004038

pH = -log(H+) = -log(0.004038) = 2.39383

pH = 2.39383

Q7.

after adding 10 mL of NaOH

find total volume first;

VT = v1+v2 = 45+20 = 65 mL

calculate mmol of each substance so we can see what is in excess

mmol of acid = M1V1 = 45*0.1 = 4.5 mmol (we previously calculated that)

mmol of base = M2V2 = 20*0.2 = 4 mmol

note htat we have excess acid so

mmol of acid LEFT = 4.5 - 4 = 0.5 mmol of acid

mmol of conjguate formed = 4 mmol

This is a buffer! since we have CHO2- and HCHO2;

apply Henderson Hasselbalch equations; for buffer

pH = pKa + log(base/acid)

for pKa; we need the next formula:

pKa = -log(Ka) = -log(1.7*10^-4) = 3.75

since this is the same solution, with same volume, no need to use Volume, compare directly with mmol:

pH = 3.75 + log(4/0.5) = 4.6530

pH = 4.6530 approx.

Q8.

In the equivalenc epoint; we have:

VT = V1+V2 = 45+22.5 = 67.5 mL

HCHO2 + NaOH --> H2O + NaCHO2

note that

NaCHO2 --> Na+ + CHO2-

This will cause the next equilibrium

CHO2- + H2O <-> HCHO2 + OH-

Whichi s basic; since OH- is present

Apply Kb value for equlibrium

Kb = [HCHO2][OH-]/[CHO2-[H2O]

Kb value can be calculated as follows:

Kw = Ka*Kb, which is the euqilibirum of water, at 25C Kw = 10^-14 always

then

10^-14 = Ka*Kb

Kb = (10^-14) / (1.7*10^-4) = 5.8823*10^-11

now...

Kb = [HCHO2][OH-]/[CHO2-[H2O]

[HCHO2] = x

[OH-] = x

[CHO2-] = M - x = 4.5/(67.5) - x = 0.066666 - x

note that M was previously calculated

[H2O] = 1, since it is liquid

then

Kb = [HCHO2][OH-]/[CHO2-[H2O]

5.8823*10^-11 = x*x/ (0.066666 - x)

solve for x

x = 1.979*10^-6

since x = [OH-]

then

[OH-] = 1.979*10^-6

we need pH so

pH = 14-pOH

pOH = -log(OH-) = -log(1.979*10^-6) = 5.703

pH = 14-5.703 = 8.297

Q9.

we have excess base, since the equivalenc epoint is 22.5 mL

then;

VT = V1+V2 = 25+45 = 70 mL

Find how many moles of base are, and reacted with the existing aicd

mmol of acid = M1V1 = 0.1*45 = 4.5 mmol of acid

mmol ofbase = M2V2 = 0.2*25 = 5 mmol of base

mmol of base left after reaction = 5-4.5 = 0.5 mmol of base left

we could calculate pOH if

[OH-] = ?

[OH-] = mmol/mL = 0.5 / 70 = 0.007142 M

and now

pOH = -log(OH-) = -log(0.007142) = 2.146180

now,

pH = 14-pOH = 14- 2.146180 =

ph = 11.85382

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