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PH Indicators / Dilution (NaOH)

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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.


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 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 ...

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.

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1. Solve the following problem related to the solubility of equilibria of some metal hydroxides in aqueous solution.

a) The solubility of Cu(OH)2 (solid)is 1.72x10^-6 gram per 100 ml of solution at 25 degrees C.
i. Write the balanced chemical equation for the dissociation of Cu(OH)2 (solid) in aqueous solution.
ii. Calculate the solubility (in moles/liter) if Cu(OH) 2 at 25 degrees C.
iii. Calculate the value of the solubility-product constant Ksp for Cu(OH)2 at 25 degrees C.
b) The value of the solubility product constant, Ksp, for Zn(OH) 2 is 7.7x10^-17 at 25 degrees C.
i. Calculate the solubility of Zn(OH)2 at 25 degrees C in a solution with a pH of 9.35
ii. At 25 degrees C, 50 mL of .1 molar Zn(NO3) 2 is mixed with 50mL of a .3 molar NaOH. Calculate the molar concentration of Zn2+ (aq) in the resulting solution once equilibrium has been established. Assume the volume are additive

2. A 30mL sample of a weak monoprotic acid was titrated with a standardized solution of NaOH. A pH meter was used to measure the pH after each increment of NaOH was added, and the curve attached was constructed

a) Explain how this curve could be useful determining the molarity of the acid
b) Explain how this curve could be used to determine the dissociation constant, K of the weak monoprotic acid.
c) If you were to repeat the titration using an indicator in the acid to signal the endpoint, which of the following indicators should you select? Why?
Methyl Red Ka= 1x10^-5
Cresol Red Ka= 1x10^-8
Alizarin yellow Ka= 1x10^-11

d) Sketch the titration curve that would result if the monoprotic acid were replaced by a strong monoprotic acid, such as HCL of the same molarity. Identify differences between this titration curve and the curve shown attached.

3. A buffer solution contains .4 mole of formic acid HCOOH, and .6 mol of sodium formate, HCOONa, in 1L of solution. The ionization constant, Ka, of formic acid is 1.8x10^-4

a) Calculate the pH of the solution
b) If 100mL of this buffer solution is diluted to a volume of 1L with pure water, the pH does not change. Discuss why the pH remains constant on dilution.
c) A 5.00mL sample of 1.00 molar HCL is added to 100mL of the original buffer solution. Calculate the [H3O-] of the resulting solution
d) An 800mL sample of 2 molar formic acid is mixed with 200mL of 4.8 molar NaOH. Calculate the [H3O-] of the resulting solution.

4. H3PO2, H3PO3, and H3PO4 are monoprotic, diprotic, and triprotic acids, respectively, and they are about equal weak acids.

HClO2, HClO3, and HClO4, are all monoprotic acids, but HClO2 is a weaker acid than HClO3, which a weaker acid than HClO4. Account for:

A) The fact that the molecules of the three phosphorus acids can provide different numbers of protons.

B) The fact that the three chlorine acids differ in strength.

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