pH change calculations?

Question

Calculate the pH change when 10 cm3 of 0.1 M NaOH is added separately to a buffer solution.

Calculate the pH change when 10 cm3 of 0.1 M NaOH is added to 120 cm3 of water.

Answer 1

Right... you have'nt provided enough information on the question involving the buffer, but the second question can be solved.

Firstly: Calculate the number of mols of NaOH present in your 10cm3 of 0.1m NaOH solution. I'll assume you can do this, using the equation n=c.v , to give you 0.001 Mols. Because NaOH is a strong alkali, it will fully dissociate into Na+ and -OH ions. You can therfore assume that your solution will contain 0.001 Mols of -OH ions.

Secondly: Water can dissociate spontaneously on its own. The concentration of H+ multiplied by the concentration of -OH will allways-no matter the PH- give you a value of 1*10^-14. This value is refered to as the water dissociation constant: it never changes. Even at low PH, when the solution contains a high conc of H+, the concentration of -OH will be smaller; when you multiply the two concentration values together- you still get 1*10^-14. And at neutral PH the concentration of H+ ions equals the concentration of -OH ions.
This information can be used to find the number of mols of -OH ions in solution in 120cm3 of water, which, can then be added onto the number of mols in the alkali sol. This will give you the total number of -OH ions in both sols. I did do this in my calculations... but to be honest the number of mols of -OH ions present in the water is sooo small in comparison to the number of ions in the acid, that I wish I had'nt wasted my time with this insignificant calculation. If your interested through I worked out that ther was 0.000, 000, 012 Mols of -OH ions in the water, which, is dwarfed by the 0.001 ion mols in the acid. So pretty much my second step is a waste of time, unless you want pin-point accuracy. I'm going to pretend that second step did'nt happen-althrough the information about the water dissociation constant will be important in my third step.

Thirdly: You have 10cm3 of 0.001 Mols of -OH ions, added to 120cm3 of water. This gives us a new volume of 130cm3, with 0.001 Mols of -OH dissolved in it. The concentration of -OH ions is therefore: 0.001Mol / 0.13dm3=0.00769 Mol dm-3.

Fourth: Using the water dissociation constant(1*10^-14), which, is allways equal to the concentration of H+ multiplied by the concentration of -OH, we can find the concentration of H+. This is required because PH is based upon the measure of H+, not -OH. By dividing the dissociation constant by the known concentration of -OH, we find the conc of H+.
(1*10^-14) / 0.00769 Mol dm-3= 1.3* 10^-12 Mol dm3.

5th: pH is related to H+ concentration by the equation
pH=-log[H+]
-log[1.3*10^-12]= pH 11.88
So finally to calculate the change in pH, its 11.88-7=4.77

Answer 2

I also have a 5500 gal tank of grease capture fabric (ninety 5% water, 5% grease) that has a ph of four.40 8. How plenty soda ash or sodium hydroxide could I could desire to dose the tank with to realize a 6.5-7.0 ph selection.

Answer 3

In both questions, we need to remember that 1cm³ = 1 mL (mililiter)

PART A:

1) In the first question, we do not know what is the composition of the buffer. The pH of any buffer solution is given by Hendersson-Hasselbach expression:

pH1 = pKa + log [Base] / [Acid]....... pH before adding NaOH

So, at the moment to add 10 mL (equal to 0.01 L) of NaOH 0.1M means that you ae adding (0.01 L)(0.1 mol/L) = 0.001 mol NaOH

In general form, you have to add this amount of moles to the concentration of [Base] in the above expression:

pH2 = pKa + log [Base+0.001] / [Acid]........pH after adding NaOH

2) So, change of pH in this case is:

pH change = pH2 - pH1


PART B:

1) To calculate the pH change, we must remember that pH water = pOH =7

2) When you add 10 cm³ of NaOH 0.1M, it means that:

(0.010 mL)(0.1mol/L) = 0.001 mol NaOH

As we know that dissotiaton of NaOH in water is:

NaOH + H20 ------- Na(+) + OH(-) + H2O

so, the 0.001 mol of NaOH are equivalent to add 0.001 moles of OH- (base) to water:

2) As pOH(water) = 7 then:
[OH-]1 = 10^(-pOH) = 1 x 10^-7 moles OH- ....before adding NaOH
[OH-]2 = 1 x 10^-7 + 0.001 moles/(0.130 L) =
= 7.692x10^-3 mol/LOH-... after adding OH-

3) Reversing: pOH2= -log [OH-] = 2.11

4) As we know: pH = 14 - pOH = 14 -2.11 = 11.88

So, the answer is:

pH change = 11.88 - 7 = 4.88

That´s it!

Good luck!