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.
Ruler is the answer of this question
A gallon of water weighs 8.34 pounds
Answer:
a H2CO3 b HCO3- and c H+ and HCO3-
Explanation:
As the pKa value of phenol is more than that of carbonic acid(H2CO3), the carbonic acid will have high Ka value than that of phenol.
The acid that contain high Ka value act as stong acid.From that point of view H2CO3 is a strong acid than phenol as the Ka value of carbonic acid is greater than that of phenol.
The conjugate base of H2CO3 is bicarbonate ion(HCO3-)
c The species that predorminates at equilibrium are H+ and HCO3-
