Question

50 points !! for this question
On the picture

Answer 1

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

Answer 2

Here's what I get

Step-by-step 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

`\begin{array}{rr}\mathbf{V} &\textbf{pH} \\0 & \mathbf{13.10} \\10 & \mathbf{12.85} \\20 & 12.62 \\30 & 12.36 \\40 & 11.95 \\50 & \mathbf{7.00} \\60 & 2.13 \\70 & \mathbf{1.88} \\80 &1.75 \\90 & 1.66 \\100 & 1.60 \\\end{array}`

7. Titration Curve

See below.