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I have a chem exam tmw and i struggle with this please help

What volume of 0.200 mol dm−3 potassium sulfate solution is required to make,
by dilution with water, 1.00 dm3
of a solution with a potassium ion concentration
of 0.100 mol dm−3?
A 100 cm3
B 250 cm3
C 400 cm3
D 500 cm3

the answer is b
idk how to right the equation ?
help
ImageUploadedByStudent Room1496881195.137210.jpg
The first equation I think


Posted from TSR Mobile
Equation is

K2SO4 + H2O ---> 2(K+) + (SO4^2-) + (HSO4^-) + (OH-)

mol of k+= 0.100*1=0.1
1:2 ratio so 0.05 mol of potassium sulphate needed
mol=conc*vol
vol= 0.05/0.20
vol=0.25 dm cubed=250 cm cubed

You dont need to know the equation, you know potassium sulphate is k2so4 and you make k+ ions, you make 2 per mole of potassium sulphate so 1:2
(edited 6 years ago)
Reply 3
Avatar for sarah5
sarah5
OP
12
Original post by glad-he-ate-her
Equation is

K2SO4 + H2O ---> 2(K+) + (SO4^2-) + (HSO4^-) + (OH-)

mol of k+= 0.100*1=0.1
1:2 ratio so 0.05 mol of potassium sulphate needed
mol=conc*vol
vol= 0.05/0.20
vol=0.25 dm cubed=250 cm cubed

You dont need to know the equation, you know potassium sulphate is k2so4 and you make k+ ions, you make 2 per mole of potassium sulphate so 1:2


thank you so much
Original post by sarah5
thank you so much


No problem

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