PH Indicators / Dilution (NaOH)
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Scenario:
You took 7 test tubes. In #1 you mixed 5ml of 0.1M NaOH and 5ml of H2O together. In the other 6 test tubes, you added 9ml of H2O. You took test tube #1 (with 5ml 0.1M NaOH and 5ml H2O) and added 1ml to test tube #2. Take test tube #2 and add 1ml to test tube #3. take test tube #3 and add 1ml to test tube #4 and so on till test tube #7. Add 2 drops (in each test tube) of Bromothymol Blue to ALL 7 test tubes.
Questions:
1. Calculate the concentrations of OH- and the pH of each solution. Since NaOH donates OH- ion for each NaOH molarity, the OH- concentration is equivalent to the NaOH molarity. The pOH is calculated as -log [OH-] and the pH is calculated as 14- pOH.
2. First, do the equation: (M con) (V con)= (M dil) (V dil)
3. Substitute known values
4. M dil = ____________ M
5. Once you find the concentration, find the pH using the formula:
pH = -log [H2O+]
6. Repeat the entire procedure using Alizain Yellow as the indicator.
7. Repeat the entire procedure using Phenolphtalein as the indicator.
8. Repeat the entire procedure using Red Cabbage Extract as the indicator.
9. Summarize your results for each pH indicator, recording both the pH values and the color of solution at each pH.
10. Estimate the useful range for each indicator, from a minimum to a maximum pH (range usually spans from 1.5 to 2pH units).
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Solution Summary
This solution of 714 words looks at calculating the concentrations of OH- and the pH of each solution with different indicators of Phenolphtalien, Alizain, Red Cabbage Extract and Bromothymol Blue. All calculations are show with full explanations.
Solution Preview
If we start with what we know, we can calculate the concentration of NaOH in tube 1 as follows.
How many moles of NaOH are in tube 1? Moles can be determined by the molarity and the volume like this (remember that 5 ml = 0.005 L):
(0.005 L)(0.1 mol/L) = 0.0005 mol of NaOH in tube 1
But what is the concentration in that tube? To find that out, we take the moles of NaOH we just calculated and divide it by the total volume, which is 10 ml:
Concentration of NaOH in tube 1 = 0.0005 mol / 0.010 L = 0.05 M
If you look at how to prepare each subsequent tube, you will notice that each one is a 1/10th dilution from the previous tube. You take 1 ml of the previous tube and put into 10 ml total of the next one. Therefore, we can easily calculate that the concentration of tube 2 must be 0.005 M. Likewise, tube 3 is 0.0005 M. And so on. You can work out the rest on your own now that you know how to do it. (Remember, we are ...
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