# Acid-Base Equilibria Help!

#1
Calculate the pH of the solution obtained by mixing 50.0 cm3
of 1 mol dm-3 sodium hydroxide solution with 49.0 cm3
of 1 mol dm-3 hydrochloric acid?

The answer is 1.96 but I’m not sure how you get it?

When you do these types of questions, what do you like out for and how do you tackle them?

When I do these, I either get them right or they go horribly horribly wrong.

Thank you so much!
0
1 month ago
#2
(Original post by Not.Vibing.)
Calculate the pH of the solution obtained by mixing 50.0 cm3
of 1 mol dm-3 sodium hydroxide solution with 49.0 cm3
of 1 mol dm-3 hydrochloric acid?

The answer is 1.96 but I’m not sure how you get it?

When you do these types of questions, what do you like out for and how do you tackle them?

When I do these, I either get them right or they go horribly horribly wrong.

Thank you so much!
I might be being stupid, but are you sure you have the concs. around the correct way? At the moment, you have more moles NaOH than HCl which would mean that 1.96 cannot be correct.
0
#3
(Original post by tony_dolby)
I might be being stupid, but are you sure you have the concs. around the correct way? At the moment, you have more moles NaOH than HCl which would mean that 1.96 cannot be correct.
I haven't changed anything about the question, just copied and pasted it straight from a past paper.

I think you use Kw to figure out what the concentration of the hydrogen ions would be?
0
1 month ago
#4
(Original post by tony_dolby)
I might be being stupid, but are you sure you have the concs. around the correct way? At the moment, you have more moles NaOH than HCl which would mean that 1.96 cannot be correct.
No I think the exact same thing the pH should be a lot higher surely
0
1 month ago
#5
Theres an excess of OH- ions, the pH has to be over 7, I got an answer of 12.00
Last edited by CreamPi; 1 month ago
2
#6
Oh, this is interesting. In my textbook - straight from the exam board - they've given the answer as 1.96.

But in the mark scheme, it's 12.

No wonder I wasn't getting it right. I'll give it another go and see what I end out with. Thanks for everyone's help!

Here's the link to the past paper:

The mark scheme:

EDIT: The question that I'm stumped with it is the first one and it's a multiple choice. But in my textbook, it's not given out as a multiple choice question but a normal one. Very weird.
Last edited by Not.Vibing.; 1 month ago
0
1 month ago
#7
(Original post by Not.Vibing.)
Oh, this is interesting. In my textbook - straight from the exam board - they've given the answer as 1.96.

But in the mark scheme, it's 12.

No wonder I wasn't getting it right. I'll give it another go and see what I end out with. Thanks for everyone's help!

Here's the link to the past paper:

The mark scheme:

EDIT: The question that I'm stumped with it is the first one and it's a multiple choice. But in my textbook, it's not given out as a multiple choice question but a normal one. Very weird.
omg 😭😭
0
#8
(Original post by aliaa03)
omg 😭😭
Honestly, I shouldn't have expected anything less. It's ccea 😭
1
1 month ago
#9
(Original post by CreamPi)
Theres an excess of OH- ions, the pH has to be over 7, I got an answer of 12.00
That's wot I got too.
0
1 month ago
#10
(Original post by Not.Vibing.)
I haven't changed anything about the question, just copied and pasted it straight from a past paper.

I think you use Kw to figure out what the concentration of the hydrogen ions would be?
Exactly.

I always set out these answers in the same way because it helps avoid careless errors.

Moles HCl = conc. x vol. = 0.0490 x 1 = 0.0490 moles
Moles NaOH = 0.0500 x 1 = 0.0500 moles

HCl + NaOH makes NaCl + H2O

Acid and base react one to one.

Work out which number of moles is greater. We will obviously have some moles NaOH left over.

Moles NaOH left = 0.0500 - 0.0490 = 0.00100 moles

Total volume of solution after addition = volume of HCl + Vol. NaOH = 0.00990 dm3

Concentration NaOH = moles/vol. = 0.001/ 0.00990 = 0.0101 mol/dm3

Now we need to use Kw

Kw = 1.00 x 10-14

[H+] = Kw/ [OH-]

[H+] = 1.00 x 10-14 / 0.0101 = 9.901 x10-13

-log 9.901 x10-13 = 12.00

pH = 12.00
1
#11
(Original post by tony_dolby)
Exactly.

I always set out these answers in the same way because it helps avoid careless errors.

Moles HCl = conc. x vol. = 0.0490 x 1 = 0.0490 moles
Moles NaOH = 0.0500 x 1 = 0.0500 moles

HCl + NaOH makes NaCl + H2O

Acid and base react one to one.

Work out which number of moles is greater. We will obviously have some moles NaOH left over.

Moles NaOH left = 0.0500 - 0.0490 = 0.00100 moles

Total volume of solution after addition = volume of HCl + Vol. NaOH = 0.00990 dm3

Concentration NaOH = moles/vol. = 0.001/ 0.00990 = 0.0101 mol/dm3

Now we need to use Kw

Kw = 1.00 x 10-14

[H+] = Kw/ [OH-]

[H+] = 1.00 x 10-14 / 0.0101 = 9.901 x10-13

-log 9.901 x10-13 = 12.00

pH = 12.00
PRSOM

Thank you so much! This is so detailed and helpful, I really do appreciate it!
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