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Eduardwww [97]
3 years ago
11

In this experiment, we will be performing a titration with a buret. place the steps in order. 1. record the ph when 0.0 ml of na

oh has been added 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 5.00 ml of naoh has been added from the buret. 3. record the ph of your partially neutralized hcl solution when 10.00ml, 15.00ml and 20.00ml of naoh has been added. 4. record the ph of your partially neutralized hcl solution when 21.00 ml, 22.00 ml, 23.00 ml and 24.00 ml of naoh has been added. 5. add naoh one drop at a time until a ph of 7.00 has been reached, then record the volume of naoh added from the buret. 6. record the ph of your basic hcl-naoh solution when 26.00 ml, 27.00 ml, 28.00 ml, 29.00 ml and 30.00 ml of naoh has been added. 7. record the ph of your basic hcl-naoh solution when 35.00 ml, 40.00 ml, 45.00ml and 50.00 ml of naoh has been added from your 50ml buret.
Chemistry
1 answer:
77julia77 [94]3 years ago
4 0

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.

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The equation representing Heisenberg's uncertainty principle follows:

\Delta x.\Delta p=\frac{h}{2\pi}

where,

\Delta x = uncertainty in position = d = 4.50\times 10^{-5}m

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h = Planck's constant = 6.627\times 10^{-34}kgm^2/s^2

Putting values in above equation, we get:

4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg}=4.424\times 10^{-5}m/s

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