# 2nd Year Chem, another pH problem! Ari!?

How much NH4CL must be added to 2.0 liters of .200M Aq Ammonia to give a solution with pH=8.2

I started the problem by subtracting the pH to find to pOH which is 5.8. then im lost.

ok. slow down. first realize that this is a buffer problem. NH4+ and NH3 are a conjugate acid base pair so its a buffer.

lets use the buffer equation from OH- concentration since the pH is basic. if pH= log [OH-] +14 then rearanging the equation gives you that 10 to the pH minus 14 power is the [OH-]. so if the pH is 8.2 the [OH-] is 1.58 x 10^ -6 M

lets set up the equation:

[OH-] = Kb x moles base / moles acid

we know that the [OH-] = 1.58 x 10^ -6 M , the Kb of ammonia is 1.8 x 10^ -5 and that there are 2.0L x 0.2M = 0.4 moles of ammonia.

1.58 x 10^ -6 = 1.8 x 10^ -5 x 0.4 moles base / X moles acid.

cross multiply:

1.58 x 10^ -6 X = (1.8 x 10^ -5)(.4)

1.58 x 10^ -6 X = 7.2 x 10^ -6

divide by 1.58 x 10^-6

X= 4.56 moles.

you need to figure out how many moles you have.

What is the reaction taking place?

(NH4+) + (OH-) -> (NH3)

Because you start with a base (ammonia) and are adding some acid. Your pH should drop from the original pH.

So, originally you had 2.0L of .200M NH3.

How many moles is that?

.200M = .200 moles / 1 L

So

2.0L * (.200 moles / 1 L) = .400 moles NH3

Your pH tells you how many moles of OH- you have.

Also, remember that this is a weak acid/base pair.

There's one more step, but I don't want to give it all to you. Hope it was enough to get you kick started.

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I started the problem by subtracting the pH to find to pOH which is 5.8. then im lost.

**Answer:**ok. slow down. first realize that this is a buffer problem. NH4+ and NH3 are a conjugate acid base pair so its a buffer.

lets use the buffer equation from OH- concentration since the pH is basic. if pH= log [OH-] +14 then rearanging the equation gives you that 10 to the pH minus 14 power is the [OH-]. so if the pH is 8.2 the [OH-] is 1.58 x 10^ -6 M

lets set up the equation:

[OH-] = Kb x moles base / moles acid

we know that the [OH-] = 1.58 x 10^ -6 M , the Kb of ammonia is 1.8 x 10^ -5 and that there are 2.0L x 0.2M = 0.4 moles of ammonia.

1.58 x 10^ -6 = 1.8 x 10^ -5 x 0.4 moles base / X moles acid.

cross multiply:

1.58 x 10^ -6 X = (1.8 x 10^ -5)(.4)

1.58 x 10^ -6 X = 7.2 x 10^ -6

divide by 1.58 x 10^-6

X= 4.56 moles.

you need to figure out how many moles you have.

What is the reaction taking place?

(NH4+) + (OH-) -> (NH3)

Because you start with a base (ammonia) and are adding some acid. Your pH should drop from the original pH.

So, originally you had 2.0L of .200M NH3.

How many moles is that?

.200M = .200 moles / 1 L

So

2.0L * (.200 moles / 1 L) = .400 moles NH3

Your pH tells you how many moles of OH- you have.

Also, remember that this is a weak acid/base pair.

There's one more step, but I don't want to give it all to you. Hope it was enough to get you kick started.

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