1. Sour pickles have a pH of about 3.00. The [OH –] in a typical sour pickle is

8. A solution of sodium hydroxide was prepared for a chemical analysis using 0.65 g in 500 mL. The hydrogen ion concentration in the solution is

Use the information to answer the next question.

Ammonia is used in industry to manufacture nitric acid, explosives, synthetic fibers, and fertilizers. In the home, ammonia is the active ingredient in many household cleaners. A solution of household ammonia is found to have a pH of 11.36.

9. In reference to the information, the hydrogen ion concentration in the solution reported to the correct certainty is

12. When the pH of a solution increases by 2.00, the [H +(aq)] increases by

13. A pH meter used to test a freshly opened carbonated soft drink gives a reading of 3.14, corresponding to a [H +(aq)] of

Use info for next question:
Amines are a class of organic compounds that can act as bases in chemical reactions.

Two examples of amines follow.

Analine, C6 H 5 NH 2(aq), Kb= 4.27>>>10–10 mol/L
Methylamine, CH 3 NH 2(aq),Kb= 3.70>>>10 – 4 mol/L

16. Write an equation for the acid/base reaction of analine with water and then calculate the Ka value of the acid that forms.

Can u please explain and help me on these..Thanks!

Five or six questions on one post is too much to handle conveniently. I would like to see how much you have done and where you are stuck instead of spending 45 minutes typing answers. So show us what you have done or tell us exactly what you don't understand about the problems. For the first one,

Remember pKw = 14 = pH + pOH.
Convert pH to pOH, then use
pOH = -log(OH^-).
I hope this helps get you started.

1st one: I get that, just how do u get OH-? How do u enter it into ur calculator. Because to figure out OH^- u must divide pOH by -log, right? How do I do that?

2nd and 3rd ones I just don't get how to do them. Please help get me started!

4th one: I believe that H+ would also increase by two if pH increased. Would this be right?

5th one:Again I have pH and H+. So my equation would be:
pH=-log(H+)
3.14= -log(H+)
My question is how I calculate it.

6th one:I don't know how to write the equation. It would be acid over base but which is the acid and which is the base?

Thats it. Thanks and sorry...

You need three equations to do all of these except the amine problem (the last one) and #8.

The first one is
pKw = 14 = pH + pOH

Second one is
pH = -log(H^+)

Third one is
pOH = -log(OH^-)


1st one: I get that, just how do u get OH-? How do u enter it into ur calculator. Because to figure out OH^- u must divide pOH by -log, right? How do I do that?
14.00 = pH + pOH
14.00 = 3.00 + pOH
14.00 - 3.00 = pOH = 11.00

pOH = -log(OH^-)
11.00 = -log(OH^-)
OR -log(OH^-) = 11.00
and changing signs on both sides gives
log(OH^-) = -11.00.
Now, TAKE THE ANTILOG of both sides. The antilog of a log, obviously, just removes the log designation. That is antilog of log(OH^-) = (OH^-)
Then for antilog of -11, punch 11 into the calculator and then the change sign key OR if there is no change sign key, punch in -11. Then look for the 10^x key and punch that. That will show 1.0 x 10^-11 as the answer.
(OH^-) = 1.0 x 10^-11 molar.


2nd and 3rd ones I just don't get how to do them. Please help get me started!
#8. The concentration of NaOH problem.
NaOH ==> Na^+ + OH^-
This is a strong electrolyte; therefore, it ionizes completely, and (OH^-) is calculated from the amount of NaOH added.
M = mols/L. How many moles did you have of NaOH? That is grams NaOH/molar mass NaOH. That many mols/L is the (OH^-). How many liters did you put it in? That is 500 mL = 0.5 L. So you can do the arithmetic. That is 0.65 NaOH/molar mass NaOH and that divided by 0.5 L = (OH^-).

The problem asks for H^+. I would change OH^- to pOH, substract from 14 to get pH, then change to (H^+)just as in the first problem above.
For the next one [ammonia problem and converting pH of 11.36 to (H^+)]. That is done as in the first one.

4th one: I believe that H+ would also increase by two if pH increased. Would this be right? Remember two things here. First, this is a log relationship; therefore, a change in pH of 1 is a change in (H^+) of 10. The second thing to remember is that the pH scale is "reversed"; i.e., when the pH goes up the (H^+) goes down. If all of this is confusing, why don't you make a couple of simple problems and solve them. Say, calculate the pH of a solution that has (H^+) = 0.1. You should find it to be pH = 1.0 (but do it to confirm that you know how to do it.) Now change that to pH = 3 and recalculate (H^+) and see how much the H^+ changed.

h one:Again I have pH and H+. So my equation would be:
pH=-log(H+)
3.14= -log(H+)
My question is how I calculate it.
You should be able to do this one from the first one above.

6th one:I don't know how to write the equation. It would be acid over base but which is the acid and which is the base?

KaKb = Kw. You know Kw, you know Kb, calculate Ka.
Analine is the base. It takes a H^+ from water to form a positive ion and leaves the OH^- alone.
I hope all of this helps. If you get stuck somewhere, please post ONE question and tell us exactly what you don't understand about it. For example, you told us you didn't know how to do log and antilogs on the calulator.

To calculate the [OH^-] in a sour pickle with a pH of 3.00, you can use the equation pOH = 14 - pH. In this case, pOH = 14 - 3.00 = 11.00.

To calculate the [OH^-] from pOH, you need to take the antilog of pOH. On most calculators, you can use the "10^x" function to find the antilog. So, if pOH = 11.00, you would calculate [OH^-] as 10^(-11), which is approximately 1.0 x 10^(-11) mol/L.

For the sodium hydroxide solution in question 8, the hydrogen ion concentration can be calculated using the equation pH = -log(H^+). In this case, you have the concentration of sodium hydroxide (0.65 g in 500 mL), so you need to convert it to molarity.

First, calculate the number of moles of sodium hydroxide:
moles = mass / molar mass = 0.65 g / 40.00 g/mol (molar mass of NaOH) = 0.01625 mol

Next, convert the volume to liters:
volume = 500 mL / 1000 mL/L = 0.5 L

Now, divide the number of moles by the volume to get the concentration:
concentration = 0.01625 mol / 0.5 L = 0.0325 mol/L

Finally, use the equation pH = -log(H^+) to find the hydrogen ion concentration:
pH = -log(H^+)
3.00 = -log(H^+)

To solve for the hydrogen ion concentration, you need to take the antilog of the pH. So, H^+ = 10^(-3.00), which is approximately 1.0 x 10^(-3) mol/L.

For the household ammonia solution with a pH of 11.36, you can again use the equation pH = -log(H^+) to find the hydrogen ion concentration. In this case, H^+ = 10^(-11.36), which is approximately 2.29 x 10^(-12) mol/L.

For the fourth question about the change in [H^+] when the pH increases by 2.00, you are correct. A change in pH of 1.00 represents a 10-fold change in [H^+], so a change in pH of 2.00 would represent a 100-fold change in [H^+].

For the equation of the acid/base reaction between aniline and water in question 6, aniline acts as the base and takes a proton (H^+) from water, forming the anilinium ion (C6H5NH3^+). The equation would be:
C6H5NH2 + H2O ⇌ C6H5NH3^+ +OH^-

To calculate the Ka value of the acid that forms, you can use the equation Ka = Kw / Kb, where Kb is the base dissociation constant. In this case, the Kb value for aniline is given as 4.27 x 10^(-10) mol/L. So, you would use the equation:
Ka = Kw / Kb = 1.0 x 10^(-14) / 4.27 x 10^(-10) = 2.34 x 10^(-5) mol/L