How many mL of acid must be added to reach the equivalence point?

(a)
NH₂OH + H⁺ → NH₂OH₂⁺
Mole ratio NH₂OH : H⁺ = 1 : 1

Amount of NH₂OH reacted = (0.150 mmol/mL) × (35.00 mL) = 5.25 mmol
Amount of HCl needed = Amount of H⁺ needed = 5.25 mmol
Volume of HCl to reach the equivalence point = (5.25 mmol) / (0.150 mmol/mL) = 35.0 mL

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(b)
The dissociation of NH₂OH :
___________ NH₂OH(aq) __ + __ H₂O(l) __ ⇌ __ NH₂OH₂⁺(aq) __ + __ OH⁻(aq) ___ Kb = 6.60 × 10⁻⁹
Initial: ______ 0.150 M _______________________ 0 M _____________ 0 M
Change: ______ -y M _______________________ +y M ____________ +y M
At eqm: ___ (0.150 - y) M ____________________ y M _____________ y M

As Kb is very small, the dissociation of NH₂OH is to a very small extent.
It is assumed that 0.150 ≫ y, and thus [NH₂OH] at eqm = (0.150 - y) M ≈ 0.150 M

Kb = [NH₂OH₂⁺] [OH⁻] / [NH₂OH]
6.60 × 10⁻⁹ = y² / 0.150
y = 3.15 × 10⁻⁵

pOH = -log[OH⁻] = -log(3.15 × 10⁻⁵) = 4.50
pH = pKw - pOH = 14.00 - 4.50 = 9.50

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(c)
NH₂OH + HCl → NH₂OH₂⁺ + Cl⁻
After 10 mL of acid has been added :
[NH₂OH] = (0.150×35.00 - 0.150×10.00)/(35.00 + 10.00) M = 0.0833 M
[NH₂OH₂⁺] = 0.150×10.00/(35.00 + 10.00) M = 0.0333 M

The dissociation of NH₂OH :
NH₂OH(aq) + H₂O(l) ⇌ NH₂OH₂⁺(aq) + OH⁻(aq) …. Kb = 6.60 × 10⁻⁹

pOH = pKb + log([NH₂OH₂⁺]/[NH₂OH]) = -log(6.60 × 10⁻⁹) + log(0.0333/0.0833) = 7.78
pH = pKw - pOH = 14.00 - 7.78 = 6.22

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(d)
At the equivalent point, NH₂OH and H⁺ completely react to give NH₂OH₂⁺ ions.
[NH₂OH₂⁺]ₒ = (0.150 M) × [35.00/(35.00 + 35.00)] = 0.0750 M

The dissociation of NH₂OH₂⁺ ions :
___________ NH₂OH₂⁺(aq) ___ ⇌ __ NH₂OH(aq) __ + __ H⁺(aq) ___ Ka = (1.00 × 10⁻¹⁴) / (6.60 × 10⁻⁹)
Initial: ______ 0.0750 M ____________ 0 M ___________ 0 M
Change: ______ -z M _______________ +z M __________ +z M
At eqm: ___ (0.0750 - z) M ___________ z M ___________ z M

As Ka is very small, the dissociation of NH₂OH₂⁺ ions is to a very small extent.
It is assumed that 0.0750 ≫ z, and thus [NH₂OH₂⁺] at eqm = (0.0750 - z) M ≈ 0.0750 M

Ka = [NH₂OH] [H⁺] / [NH₂OH₂⁺]
(1.00 × 10⁻¹⁴) / (6.60 × 10⁻⁹) = z² / 0.0750
z = 3.37 × 10⁻⁴

pH = -log[H⁺] = -log(3.37 × 10⁻⁴) = 3.47

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(e)
When 40.00 mL HCl has been added, HCl is in excess.
Excess [H⁺] = (0.150×40.00 - 0.150×35.00)/(40.00 + 35.00) M = 0.01 M
As NH₂OH₂⁺ is a weak acid and due to the common ion effect in the presence of H⁺, the dissociation of NH₂OH₂⁺ is negligible.

pH = -log[H⁺] = -log(0.01) = 2.00