Friday

November 27, 2015
Posted by **!!!** on Sunday, June 18, 2006 at 7:27pm.

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 dont 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 dont 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...

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 = 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 dont get how to do them. Please help get me started!

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.

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

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.

6th one:I dont know how to write the equation. It would be acid over base but which is the acid and which is the base?

Analine is the base. It takes a H^+ from water to form a positive ion and leaves the OH^- alone.