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Titration of weak acid with strong base.....

25.0 mL of acid with 10.0 mL of titrant. Concentrations are reduced to:
25.0 mL HNO2 x (0.128 mol HNO2 / 1L) / 35.0 mL = 0.09143 M HNO2
10.0 mL OH- x (0.122 mol OH- / 1L) / 35.0 mL = 0.03486 M OH-

HNO2(aq) + OH^-(aq) --> NO2^-(aq) + HOH(l)
0.09143M ...0.03486M.... ...................... .................. before
0.05657M .....0 ............... 0.03486M ........................ after

HNO2 <==> H+ + NO2^- ............. Ka = 7.2x10^-4
0.05657....... 0 .....0.03486 .......... initial
-x ............... +x .... +x .................. change
0.05657-x..... x ....0.03486+x

Ka = [H+][NO2^-] / [HNO2]
7.2x10^-4 = x(0.03486+x) / (0.05657-x)
x = 0.00111
[H+] = 0.00111M
pH = -log[H+] = 2.95 ............ to two significant digits*

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25.0 mL of acid with 10.0 mL of titrant. Concentrations are reduced to:
25.0 mL HNO2 x (0.128 mol HNO2 / 1L) / 45.0 mL = 0.07111 M HNO2
20.0 mL OH- x (0.122 mol OH- / 1L) / 45.0 mL = 0.05422 M OH-

HNO2(aq) + OH^-(aq) --> NO2^-(aq) + HOH(l)
0.07111M ...0.05422M.... ...................... .................. before
0.01689M .....0 ............... 0.05422M ........................ after

HNO2 <==> H+ + NO2^- ............. Ka = 7.2x10^-4
0.01689....... 0 .....0.05422 .......... initial
-x ............... +x .... +x .................. change
0.01689-x..... x ....0.05422+x

Ka = [H+][NO2^-] / [HNO2]
7.2x10^-4 = x(0.05422+x) / (0.01689-x)
x = 8.895x10^-4
[H+] = 8.895x10^-4M
pH = -log[H+] = 3.05............ to two significant digits*

* When taking the log of a number, only the digits to the right of the decimal reflect the precision of the original number. Based on the precision of Ka (2 significant digits), the final answer is expressed to two significant digits.

Part 1 :

Initial number of moles of HNO₂ = (0.128 mol/L) × (25.00/1000 L) = 0.00320 mol
Initial number of moles of OH⁻ = (0.122 mol/L) × (10.00/1000 L) = 0.00122 mol
Volume of the solution = (25.00 + 10.00) mL = 35.00 mL = 0.03500 L

HNO₂ + OH⁻ → H₂O + NO₂⁻
Mole ratio HNO₂ : OH⁻ = 1 : 1
HNO₂ is in excess.
[HNO₂] = {(0.00320 - 0.00122) mol} / (0.03500 L) = 0.0566 M

Consider the dissociation of HNO₂ :
____________ HNO₂ (aq) __ ⇌ __ NO₂⁻(aq) __ + __ H⁺(aq) ___ Ka = 7.2 × 10⁻⁴
Initial: ______ 0.0566 M ________ 0 M __________ 0 M
Change: ______ -y M __________ +y M _________ +y M
At eqm: ___ (0.0566 - y) M ______ y M __________ y M

At equilibrium :
Ka = [NO₂⁻] [H⁺] / [HNO₂]
7.2 × 10⁻⁴ = y² / (0.0566 - y)
y² + (7.2 × 10⁻⁴)y - (4.0752 × 10⁻⁵) = 0
y = {-(7.2 × 10⁻⁴) ± √[(7.2 × 10⁻⁴)² + 4×(4.0752 × 10⁻⁵)]} / 2
y = 0.00603 or y = -0.00675 (rejected)

pH = -log(0.00603) = 2.22

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Part 2 :

Initial number of moles of HNO₂ = (0.128 mol/L) × (25.00/1000 L) = 0.00320 mol
Initial number of moles of OH⁻ = (0.122 mol/L) × (20.00/1000 L) = 0.00244 mol
Volume of the solution = (25.00 + 20.00) mL = 45.00 mL = 0.04500 L

HNO₂ + OH⁻ → H₂O + NO₂⁻
Mole ratio HNO₂ : OH⁻ = 1 : 1
HNO₂ is in excess.
[HNO₂] = {(0.00320 - 0.00244) mol} / (0.04500 L) = 0.0169 M

Consider the dissociation of HNO₂ :
____________ HNO₂ (aq) __ ⇌ __ NO₂⁻(aq) __ + __ H⁺(aq) ___ Ka = 7.2 × 10⁻⁴
Initial: ______ 0.0169 M ________ 0 M __________ 0 M
Change: ______ -z M __________ +z M _________ +z M
At eqm: ___ (0.0169 - z) M ______ z M __________ z M

At equilibrium :
Ka = [NO₂⁻] [H⁺] / [HNO₂]
7.2 × 10⁻⁴ = z² / (0.0169 - y)
z² + (7.2 × 10⁻⁴)z - (1.2168 × 10⁻⁵) = 0
z = {-(7.2 × 10⁻⁴) ± √[(7.2 × 10⁻⁴)² + 4×(1.2168 × 10⁻⁵)]} / 2
z = 0.00315 or z = -0.00387 (rejected)

pH = -log(0.00315) = 2.50