# general chem.

posted by
**!!!**
.

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.

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.

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.

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.*