Answer:
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
[tex]\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}[/tex]
7. Titration Curve
See below.
