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.
Answer:
0.159 M
Explanation:
convert from mL to L then use the equation:
M1V1 = M2V2
rearrange to find M2
Answer:
Pag-alaga ng kalikasan, pagpapakita ng mga tanawin o kagandahan ng Pilipinas
Potassium oxide has the antifluorite structure. The antifluorite structure have compounds with the stoichiometry X₂Y, where X is the cation and Y is the anion. In the antifluorite structure <span>positions of the </span>cations<span> and </span>anions<span> are reversed relative to their positions in calcium fluoride.</span>
Potassium ions coordinated to 4 oxide ions, <span>potassium ions are all in the tetrahedral holes.</span>
