Question

Could someone help with a pH question. My book is really confusing with how its explaining how to do it. Please help!

Answer 1

So the first question asks this: Find the number of moles of HA in the 50 ml of 0.01 M acetic acid. I got 0.005M/L as an answer.

The second question asks this: Find the number of moles of NaOH in 1 ml of 0.20 NaOH. I got 0.02 m/L as an answer.

Answer 2

The third question asks to find the pH using the henderson hasselbalch equation after the addition of 1 ml of 0.20 M NaOH. I got 5.36 as an answer.

Answer 3

The fourth question which im having problems with asks this; Find the pH after the addition of 2,3,4,5....24 mls of NaOH. Each time subtracting the number of moles of NaOH added from the number of HA moles you started with and adding it to the number of A-. Once you reached 24, find ph for 24.2 etc. What is the volume that the ph will be equal to the pK

Answer 4

I don't understand your answer to the first part. what is M/L ?

Answer 5

moles per liter

Answer 6

but the question asks
Find the number of moles of HA in the 50 ml

so the answer should be moles? (not moles per liter)

Answer 7

i just used the moles=concentration x volume in liters equation to find it

Answer 8

I assume you do
0.01 moles/liter * 0.050 liters = 0.0005 moles

Answer 9

ugh sorry...it should be in moles...I forgot that M= moles/liter so the liters cancels out to give me the unit of moles

Answer 10

you also seem to be off by a decimal place ?

Answer 11

ugh sorry, its been so long since i did these types of calculations....im gonna try it again and see what i get. I will post what i got on here after my second try .

Answer 12

How do we do part 3 ?

Answer 13

ok. so I made a mistake in typing. It should be 0.10 m instead of 0.01 m. And for part 3 i use the henderson hasselbalch equation after finding the moles of each compound and finding the amount of base made

Answer 14

You wrote
***The second question asks this: Find the number of moles of NaOH in 1 ml of 0.20 NaOH. I got 0.02 m/L as an answer. ***

Is that correct? isn't it 0.001L * 0.2 moles/L ?

Answer 15

yes

Answer 16

how do you get 0.02 from 0.001*0.2 = 0.0002

Answer 17

calculation error. I fixed it in my new calc

Answer 18

what are the answers for parts 1 through 3 ?

Answer 19

for part 1 i got 0.005 moles and 0.0002 moles....ph of 3.38

Answer 20

for part 3, is it
-log(1.8E-5) + log(.0002/0.005)
4.74 - 1.40 = 3.35
?

Answer 21

its ph= 4.76 (given) + log (0.0002/0.0048)

Answer 22

oh. so its log(0.0002/(0.005-0.0002) ) + 4.76

ok that makes sense (matches up with part 4)

Answer 23

For part 4, you make a table
2,3,4,5....24 mls of NaOH.
ml of NaOH moles of NaOH HA
0 0 0.005
1 .0002 0.0048
2 .0004 0.0044
3 .0006 0.0038
and so on
calculate the pH as you go

Answer 24

I'm assuming we start by adding 2 ml of NaOH to the solution from part 3 (that already has 1 ml of NaOH)

Answer 25

no, that does not make sense. Maybe they mean, "add 1 ml" at each step, so that the total is 2,3,4,5... 24 ml. Like this
ml of NaOH moles of NaOH HA
0 0 0.005
1 .0002 0.0048
2 .0002 0.0046
3 .0002 0.0044
and so on ...

Answer 26

so the pH calculation for 3 ml would be
log(0.0006/(0.005-0.0006) ) + 4.76 = 3.89

Answer 27

Or maybe the table should look like this
ml of NaOH Total moles of NaOH HA
0 0 0.005
1 .0002 0.0048
2 .0004 0.0046
3 .0006 0.0044

Answer 28

hold on...im making the table myself

Answer 29

yes the table is what i got

Answer 30

Once you reached 24, find ph for 24.2 etc.

That sounds like once you get to 24 mil
then start calculating the pH ? I don't know what find pH for 24.2 etc
means

Answer 31

i have to find the ph at each new addition of base . so i have to find it at 2, then 3, then 4, then 5.....then at 24.2, 24,4, 24.6

Answer 32

are you sure you have to find the pH for the entries from 2 unto 24 , or just after 24 ?

But I assume you can finish this? I have to leave

Answer 33

all. and yeah im ok doing it