Please help, I'm confused and it's due very soon!
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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
(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.
