Question

Can someone solve Thierry two problems

Answer 1

Answer:

5.00 × 10⁻⁴ mol HCl; 100 mL NaOH

Step-by-step explanation:

3. Moles of HCl

Data:

V(HCl) = 500 mL

    pH = 3

Calculations:

            pH = 3

      [H₃O⁺] = 10⁻³ mol·L⁻¹

              V = 500 mL × (1 L/1000 mL) = 0.500 L

Moles HCl = volume × concentration

                 = 0.500 L × (10⁻³ mol/1 L)

                 = 5.00 × 10⁻⁴ mol

4. Volume of NaOH

NaOH + HCl ⟶ NaCl + H₂O

Data:

[NaOH] = 0.01 mol·L⁻¹

   [HCl] = 0.02 mol·L⁻¹

V(HCl) = 50 mL

Calculations:

    Moles HCl = 0.050 L × (0.02 mol/1 L)

                      = 1.0 × 10⁻³ mol

Moles NaOH = 1.0 × 10⁻³ mol HCl × (1 mol NaOH/1 mol HCl)

                      = 1.0 × 10⁻³ mol NaOH

      V(NaOH) = 1.0 × 10⁻³ mol NaOH × (1 L NaOH/0.01 mol NaOH)

                      = 0.1 L NaOH     Convert to millilitres

                      = 100 mL NaOH