50 points !! for this question
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Here's what I get  

Explanation:

1. At 0 mL HCl

pOH = -log[OH⁻] = -log(0.125) = 0.90

pH = 14.00 - pOH = 14.00 - 0.90 = 13.10

2. At 10 mL HCl

Initial moles NaOH = 25.0 mL × 0.125 mmol/mL = 3.125 mmol NaOH

Moles HCl added = 10 mmol × 0.0625 mmol/L = 0.625 mmol HCl

Moles NaOH remaining = (3.125 - 0.625) mmol = 2.50 mmol NaOH

Total volume = (25.0 + 10) mL= 35.0 mL

[OH⁻] =2.50/35 = 0.071 43 mol·L⁻¹

pOH = 1.15

pH = 12.85

3. At 50 mL HCl

Initial moles NaOH = 3.125 mmol NaOH

Moles HCl added =3.125 mmol HCl

Moles NaOH remaining = 0

We are at the equivalence point.

pH = 7.00

4. At 70 mL HCl

Initial moles NaOH = 3.125 mmol NaOH

Moles HCl added =4.375 mmol HCl

Excess moles HCl = 1.25 mmol HCl

Total volume = 95 mL

[H⁺] = 0.013 16 mol·L⁻¹

pH = 1.88

5. Table

7. Titration Curve

See below.

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