Order.
Hope this helps! :)
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.
Answer:
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- <u><em>pOH = 0.36</em></u>
Explanation:
Both <em>potassium hydroxide</em> and <em>lithium hydroxide </em>solutions are strong bases, so you assume 100% dissociation.
<u>1. Potassium hydroxide solution, KOH</u>
- Volume, V = 304 mL = 0.304 liter
- number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol
- 1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944
<u>2. LIthium hydroxide, LiOH</u>
- Volume, V = 341 mL = 0.341 liter
- number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol
- 1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391
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<u>3. Resulting solution</u>
- Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol
- Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter
- Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M
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<u>4. </u><em><u>pOH</u></em>
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