I am guessing that your solutions of HCl and of NaOH have approximately the same concentrations. Then the equivalence point will occur at pH 7 near 25 mL NaOH.
The steps are already in the correct order.
1. Record the pH when you have added 0 mL of NaOH to your beaker containing 25 mL of HCl and 25 mL of deionized water.
2. Record the pH of your partially neutralized HCl solution when you have added 5.00 mL of NaOH from the buret.
3. Record the pH of your partially neutralized HCl solution when you have added 10.00 mL, 15.00 mL and 20.00 mL of NaOH.
4. Record the NaOH of your partially neutralized HCl solution when you have added 21.00 mL, 22.00 mL, 23.00 mL and 24.00 mL of NaOH.
5. Add NaOH one drop at a time until you reach a pH of 7.00, then record the volume of NaOH added from the buret ( at about 25 mL).
6. Record the pH of your basic HCl-NaOH solution when you have added 26.00 mL, 27.00 mL, 28.00 mL, 29.00 mL and 30.00 mL of NaOH.
7. Record the pH of your basic HCl-NaOH solution when you have added 35.00 mL, 40.00 mL, 45.00 mL and 50.00 mL of NaOH from your 50mL buret.
Actions form positive ions while anions forms negative
The correct answer is letter C
Answer:
in prism
it's from the rectangular reflecting surface
Answer:
5.702 mol K₂SO₄
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Compounds
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 993.6 g K₂SO₄
[Solve] moles K₂SO₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of K: 39.10 g/mol
[PT] Molar Mass of S: 32.07 g/mol
[PT] Molar mass of O: 16.00 g/mol
Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 4 sig figs.</em>
5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄
