A-level chem calc Q

Watch this thread
sarabham2001
Badges: 2
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 2 years ago
#1
A 50.0 cm3
sample of solution X was added to 50 cm3
of dilute sulfuric acid and
made up to 250 cm3
of solution in a volumetric flask.
A 25.0 cm3
sample of this solution from the volumetric flask was titrated with a
0.0205 mol dm−3 solution of KMnO4
At the end point of the reaction, the volume of KMnO4 solution added was 18.70 cm3.
i) Write an equation for the reaction between iron(II) ions and manganate(VII) ions.
Use this equation and the information given to calculate the concentration of
iron(II) ions in the original solution X.

ANSWER:
MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Moles KMnO4 = 18.7 × 0.0205 / 1000 = (3.8335 × 10−4)
Moles Fe2+ = 5 × 3.8335 × 10−4 = 1.91675 × 10−3
Mark for M2 × 5
Moles Fe2+ in 250 cm3
= 10 × 1.91675 × 10−3 = 0.0191675 moles in 50
cm3

Original conc Fe2+ = 0.0191675 × 1000 / 50 = 0.383 mol dm−3

Why have they multiplied by 10? Is it just to get from 25 to 250 (that seems too simple)?
0
reply
David Getling
Badges: 19
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 2 years ago
#2
(Original post by sarabham2001)
Why have they multiplied by 10? Is it just to get from 25 to 250 (that seems too simple)?
Yes it is that simple. It's also something that students forget to do and lose marks for.
1
reply
Sinthuja Vijay
Badges: 15
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 1 year ago
#3
(Original post by sarabham2001)
A 50.0 cm3
sample of solution X was added to 50 cm3
of dilute sulfuric acid and
made up to 250 cm3
of solution in a volumetric flask.
A 25.0 cm3
sample of this solution from the volumetric flask was titrated with a
0.0205 mol dm−3 solution of KMnO4
At the end point of the reaction, the volume of KMnO4 solution added was 18.70 cm3.
i) Write an equation for the reaction between iron(II) ions and manganate(VII) ions.
Use this equation and the information given to calculate the concentration of
iron(II) ions in the original solution X.

ANSWER:
MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Moles KMnO4 = 18.7 × 0.0205 / 1000 = (3.8335 × 10−4)
Moles Fe2+ = 5 × 3.8335 × 10−4 = 1.91675 × 10−3
Mark for M2 × 5
Moles Fe2+ in 250 cm3
= 10 × 1.91675 × 10−3 = 0.0191675 moles in 50
cm3

Original conc Fe2+ = 0.0191675 × 1000 / 50 = 0.383 mol dm−3

Why have they multiplied by 10? Is it just to get from 25 to 250 (that seems too simple)?
does anyone know why you divide by 1000?
0
reply
cefox
Badges: 9
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 1 year ago
#4
(Original post by Sinthuja Vijay)
does anyone know why you divide by 1000?
This is the conversion from cm3 to dm3
Will come up in most titration calculations as concentrations are generally in mol dm-3 but titration results will be in cm3....so no. of moles = conc (in mol dm-3) x volume (in dm3)
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest

How did The Student Room help you with your university application?

Talking to current university students (12)
17.65%
Talking to peers going through the same thing (24)
35.29%
Speaking to student ambassadors from the universities (5)
7.35%
Speaking to staff members from universities (1)
1.47%
Using the personal statement builder, library or helper service (7)
10.29%
Reading articles about what steps to take (14)
20.59%
Learning about/speaking to Student Finance England (2)
2.94%
Something else (tell us in the thread) (3)
4.41%

Watched Threads

View All