Question

2. The weak base, hydroxylamine (HONH2) is generally stable in solution but can easily explode if dried into its solid state. The pKa of its conjugate acid is 5.96 (Harris, Appendix G). A 50.0 mL solution containing hydroxylamine is titrated to the endpoint with 32.84 mL of 0.0568 M HCl.

a) Write out the chemical equation for the aqueous equilibrium of this aqueous monoprotic base before the addition of any acid. Also write out the also the mathematical equilibrium expression. What fraction of hydroylamine is dissociated into ions, initially?

b) Write the expression for the relevant equilibrium process occurring following complete neutralization (to the equivalence point) described above. What fraction of the conjugate acid is dissociated into ions at this point?

c) What acid/base indicator is best suited to the pH of this titration endpoint? Support your choice quantitatively, and use the textbook table of indicators in Chapter 10 of Harris 8th Edition of Quantitative Chemical Analysis.

d) Since this is the titration of a very weak base, identify a possible analytical approach you could use to improve the quantification of hydroxylamine; in other words, when the endpoint isn’t very clear, what is an example of a technique used to titrate very weak acids or bases?

Answer 1

Q2.

a)

before any addition of acid:

HONH2(aq) + H2O(l) --> HONH3+(aq) + OH-(aq)

which forms OH- i.e. base!

for the equilibrium:

conjugte acid is HONH3+ and has a pKa = 5.96

meaning, pKb = 14-pKa = 14-5.96 = 8.04

the pKb of HONH2 = 8.04

HONH2 + H2O --> HONH3+ + OH- Kb = 10^-pOH = 10^-8.04 =9.120*10^-9

Kb = [HONH3+][OH-]/[HONH2 ]

%fraction of BH+ = [OH-]/M*100%

calculate

Kb = [HONH3+][OH-]/[HONH2 ]

9.120*10^-9 = (x*x)/(M-x)

find initial concentration

mmol of acid = MV = 32.84*0.0568 = 1.865312

mmol of base = 1.865312 (according to 1:1 ratio)

[Base] = mmol of base / mL = 1.865312/(50) = 0.03730 M

so

9.120*10^-9 = (x*x)/(M-x)

substitute M:

9.120*10^-9 = (x*x)/(0.03730 -x)

solve for x

x = 1.843*10^-5

[OH-] = x= 1.843*10^-5

so

%ion = [OH-]/M*100% = (1.843*10^-5) / (0.03730 )*100% = 0.0494 % is ionized

b)

in neutralization

HONH3+ will form:

HONH3+(aq) + H2O(l) <--> HONH2(aq) + H3O+(aq)

%fraction of conjgute acid (HONH3+) dissociated

Ka = [HONH2][H3O+]/[HONH3+]

10^-pKa = x*x/(Mnew-x)

note that Mnew is in equilibrium...

Vtotal = Vacid + Vbase = 50 + 32.84 = 82.84

recall that we had = 1.865312 mmol of base so mmol of conjugate = 1:1 ratio = 1.865312

[conjugate ] = 1.865312/82.84 = 0.022517 M

10^-5.96= x*x/(0.022517 -x)

x = [H3O+] = 1.565*10^-4

so

%ion = [H+]/ M *100% = (1.565*10^-4)/0.022517 *100 = 0.69503%

c)

the pH will be

pH = -log(H+) = -log(1.565*10^-4) = 3.80

try methyl orange since range is

below pH 3.1 above pH 4.4

3.1 ⇌ 4.4

d)

When we have very dilute weak acids --> use conductometric titration, i.e. conductivity as a new standard (vs. pH)