Gen Chem question

[SayAahh]

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Hey guys, maybe I'm overthinking this but anyhow, I thought I'd see how you guys went around solving it;

What's molarity of K2SO4 in aq solution made by adding 112g solid KOH to 500mL of 0.50M H2SO4(aq) ?

Thanks in advance!

 

[drgreen]

firstly molarity is defined as moles of solute per litre solution

2KOH + H2SO4 -> K2SO4 + 2H2O

Moles of solute (K2SO4):

n=m/M
=112/(39+16+1)
=2 moles KOH

n=0.5mol/L*.5Litres=0.25 mol H2SO4

since the molar ratio of KOH to H2SO4 is 2:1, our limiting reagent is H2SO4 because we have less than half the moles of KOH of H2SO4.

Since we have 0.25 mol H2SO4, using the balanced equation we find that we get 0.25 mol K2SO4. (1:1 molar ratio H2SO4:K2SO4)

Litres solution:


We have a 500 mL or .5 L solution as seen in the problem

Final Calculation:


Molarity K2SO4 =mol K2SO4/Litres Solution
=.25molK2SO4/.5Litres solution
=.125 (mol/L)

I'm pretty sure that's how it's done, let me know what you think

 

[ygenome]

firstly molarity is defined as moles of solute per litre solution

2KOH + H2SO4 -> K2SO4 + 2H2O

Moles of solute (K2SO4):

n=m/M
=112/(39+16+1)
=2 moles KOH

n=0.5mol/L*.5Litres=0.25 mol H2SO4

since the molar ratio of KOH to H2SO4 is 2:1, our limiting reagent is H2SO4 because we have less than half the moles of KOH of H2SO4.

Since we have 0.25 mol H2SO4, using the balanced equation we find that we get 0.25 mol K2SO4. (1:1 molar ratio H2SO4:K2SO4)

Litres solution:


We have a 500 mL or .5 L solution as seen in the problem

Final Calculation:


Molarity K2SO4 =mol K2SO4/Litres Solution
=.25molK2SO4/.5Litres solution
=.125 (mol/L)

I'm pretty sure that's how it's done, let me know what you think

I think your final calculation is wrong. It should be 0.5 mol/L instead of 0.125 mol/L.

 

[SayAahh]

I get it! Thank you 🙂