Question

A student was given the task of titrating a 20.mL sample of 0.10MHCl(aq) with 0.10MNaOH(aq) . The HCl(aq) was placed in an Erlenmeyer flask. An equation for the reaction that occurs during the titration is given above.
The box below to the left represents ions in a certain volume of 0.10MHCl(aq) . In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq)

Answer 1

Answer:

Here's what I get  

Step-by-step Explanation

(a) Effect of dilution

There will be no effect on the volume of NaOH needed.

The amount of HCl will be halved, so the amount of NaOH will be halved.

However, the concentration of NaOH is also halved, so you will need twice the volume.

You will be back to the same volume as before dilution.

(b) Net ionic equation

Molecular: HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + Cl⁻(aq) + H₂O(l)

Net ionic: H⁺(aq) + OH⁻(aq) ⟶ H₂O(l)

(c) Proton acceptor

H⁺ is the proton. OH⁻ accepts the proton and forms water.

(d) Moles of HCl

$\text{Moles of HCl} = \text{20. mL HCl} \times \dfrac{\text{0.10 mmol HCl}}{\text{1 mL HCl}} = \text{2.0 mmol HCl} = \textbf{0.0020 mol HCl}$

(e) Equivalence point

The equivalence point is the point at which the titration curve intersects the pH 7 line.

(f) Schematic representation

Assume the box for 0.10 mol·L⁻¹ HCl contains four black dots (H⁺) and four open circles (Cl⁻).

The 0.20 mol·L⁻¹ solution is twice as concentrated.

It will contain eight black dots and eight open circles.