Concentration Calculation

13

I have to do some concentration questions for summer homework, but I can't remember how to do them. Could someone help me with the first one please?:

25cm³ of a solution of 0.1 M NaOH reacts with 5.0 cm³ of a solution of hydrochloric acid. What is the morality of the acid?

Hopefully I'll be able to do the rest after someone explains the first! Thanks in advance.

Reply 1

Aired

1

VeerHS

I have to do some concentration questions for summer homework, but I can't remember how to do them. Could someone help me with the first one please?:

25cm³ of a solution of 0.1 M NaOH reacts with 5.0 cm³ of a solution of hydrochloric acid. What is the morality of the acid?

Hopefully I'll be able to do the rest after someone explains the first! Thanks in advance.

NaOH + HCl --> NaCl + H₂O

25cm³ of 0.1 molar NaOH = (25/1000)x0.1 moles
= 0.0025 moles NaOH

Therefore, there are 0.0025 moles HCl (1:1 molar ratio)

0.0025 moles HCl in 5.0cm³
Therefore, in 1dm³ there are (1000/5) x 0.0025 moles

= 0.5M HCl.

Reply 2

Kubed

OP

13

Thanks Aired!

Reply 3

Aired

1

No problem :smile:

Reply 4

Kicky

2

Just remember that M1 x V1 = M2 x V2 (1st molarity x 1st volume =2nd molarity x 2nd volume) and re arrange the equation to find the volume or concentration that you want.

:smile:

EDIT: ignore this-see Aired's post

Reply 5

Aired

1

Kicky

Just remember that M1 x V1 = M2 x V2 (1st molarity x 1st volume =2nd molarity x 2nd volume) and re arrange the equation to find the volume or concentration that you want.

:smile:

The equation is only correct when the molar ratios are the same, I.e. in the question that VeerHS posted.

Take the reaction between calcium hydroxide and hydrochloric acid as an example:

Ca(OH)₂ + 2HCl --> CaCl₂ + 2H₂O

25cm³ of 1 molar Ca(OH)₂ (= 0.025 moles)
100cm³ of 0.5 molar HCl (=0.05 moles)

If the equation you posted was always correct, then this should work...

(1 x 25) = (100 x 0.5)
25 = 50 - This is obviously incorrect.

Reply 6

Kicky

2

oops

I always knew that my chemistry teacher lied :p:

Reply 7

Aired

1

Kicky

oops

I always knew that my chemistry teacher lied :p:

It was probably good enough to use for GCSE calculations anyway :P

Reply 8

charco

Study Forum Helper

18

VeerHS

I have to do some concentration questions for summer homework, but I can't remember how to do them. Could someone help me with the first one please?:

25cm³ of a solution of 0.1 M NaOH reacts with 5.0 cm³ of a solution of hydrochloric acid. What is the morality of the acid?

Hopefully I'll be able to do the rest after someone explains the first! Thanks in advance.

Acids have no morality - however this does not mean that they are immoral they are inanimate and consequently incapable of logical thought.

Reply 9

Aired

1

charco

Acids have no morality - however this does not mean that they are immoral they are inanimate and consequently incapable of logical thought.

lol...

Reply 10

Kubed

OP

13

LOL Charco...Thanks everyone for help

Reply 11

mahesh

M1 * V1 = M2 * V2 is not always true.... Instead use N1 * V1 = N2 * V2

where N represents the normality... Now how to calculate normality from molarity is the question...
Remember.... Normality = Molarity/(acidity of base or basicity of acid)

Acidity of base = No. of replacable OH- ions.
basicity of acid = No of replacable H+ ions.

For NAOH.... Normality=Molarity/1 (since there is only one replacable OH- ion)
For HCL..... Normality= Molarity/1 ( Since there is only one replacable H+ ion)
For H2SO4... Normality= Molarity/2 ( 2 replacable H+ ions)
For Ca2SO4 Normality= Molarity/2 ( 2 replacable OH- ions)

If u remember this method..u will never get confused.
Think this will help u....

Reply 12

Aired

1

Molarity = Moles/Volume (dm³) is the only real formula you need for this type of question...

Reply 13

oxymoron

7

Aired

Molarity = Moles/Volume (dm³) is the only real formula you need for this type of question...

Which is easily remembered if you recall the units of concentration (molarity) are mol/dm3

Reply 14

Kubed

OP

13

I can't do another one...Can anyone help? In a titration of 25 cm3 ammonia solution reacts with 33.30 cm3 of 0.1 M HCL. What is the concentration of the ammonia solution in grams dm-3
?
Thanks

Reply 15

oxymoron

7

VeerHS

I can't do another one...Can anyone help? In a titration of 25 cm3 ammonia solution reacts with 33.30 cm3 of 0.1 M HCL. What is the concentration of the ammonia solution in grams dm-3
?
Thanks

This is a general scheme that should help you answer many of these types of questions ...

1. Write down the reaction equation
2. Work out how many moles of HCl reacted
3. From (1) and (2) calculate how many moles of ammonia reacted and hence are in 25cm3
4. Scale this to a concentration in mol/dm3
5. Convert this to g/dm3 using the molar mass of ammonia

If you tell me which step you are stuck on then I will give you more detailed help.

Reply 16

Kubed

OP

13

Thanks...I'm not sure how to do step 5 though

Reply 17

oxymoron

7

VeerHS

Thanks...I'm not sure how to do step 5 though

OK ... if, for example, I find the concentration is 0.5mol/dm3 (of ammonia) ...

then all you need to do is convert that 0.5 mol into grams using the molar mass of NH3 which is 17.

so mass = 17 x 0.5 = 8.5 g

so this means you have 0.5 moles or 8.5g in 1dm3

concentration = 0.5mol/dm3 or 8.5g/dm3

Reply 18

Borek

4

In general - for concentration conversions - try

http://www.chembuddy.com/?left=concentration&right=toc

Best,
Borek
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Concentration Calculation

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