titration help !!!!!!!!!!!!

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ogbroskicat98765
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#1
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#1
A solution of potassium hydroxide (KOH) was titrated against a solution of hydrochloric acid. It took 35cm3 of the hydrochloric acid to completely neutralise 50cm3 of potassium hydroxide. Work out the concentration of the potassium hydroxide solution in mol/dm3 if the concentration of the acid was 2mol/dm³. Round your answer to 1 d.p.
answer must be in mol/dm3
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Pigster
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(Original post by ogbroskicat98765)
A solution of potassium hydroxide (KOH) was titrated against a solution of hydrochloric acid. It took 35cm3 of the hydrochloric acid to completely neutralise 50cm3 of potassium hydroxide. Work out the concentration of the potassium hydroxide solution in mol/dm3 if the concentration of the acid was 2mol/dm³. Round your answer to 1 d.p.
answer must be in mol/dm3
What have you tried, so far?
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ogbroskicat98765
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(Original post by Pigster)
What have you tried, so far?
honestly, I haven't tried anything yet
I was hoping for advice on how to do it
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Pigster
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(Original post by dianagrace)
first you'd need to write the balanced symbol equation of the reaction between KOH and HCL which would be:
KOH+HCL-> KCL+H20
you know that 35cm^3 of 2 mol/dm^3 HCL has been used, and you need to work out the moles of the HCL from this info using the equation:
concentration(mol/dm^3 )=moles(mol)/volume(dm^3)
rearranging the equation to make moles the subject, you'd get: moles=concentration x volume
plugging in the info, you'd get moles of HCL= 35/1000 dm^3 x 2 moldm^3= 0.07 mol
because there's a 1:1 ratio of HCL:KOH, the moles of KOH would also equal 0.07 mol
now using the con equation, you'd get the concentration of KOH= 0.07mol/(50/1000) dm^3= 1.4 moldm^3
hope this helps and sorry if any of my calculations are wrong! you may want to double check all the calculations I've made in case of silly mistakes
You are correct.

But I recommend not doing other people's work for them and instead try to guide people through a problem - people learn much better that way.
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charco
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(Original post by dianagrace)
first you'd need to write the balanced symbol equation of the reaction between KOH and HCL which would be:
KOH+HCL-> KCL+H20
you know that 35cm^3 of 2 mol/dm^3 HCL has been used, and you need to work out the moles of the HCL from this info using the equation:
concentration(mol/dm^3 )=moles(mol)/volume(dm^3)
rearranging the equation to make moles the subject, you'd get: moles=concentration x volume
plugging in the info, you'd get moles of HCL= 35/1000 dm^3 x 2 moldm^3= 0.07 mol
because there's a 1:1 ratio of HCL:KOH, the moles of KOH would also equal 0.07 mol
now using the con equation, you'd get the concentration of KOH= 0.07mol/(50/1000) dm^3= 1.4 moldm^3
hope this helps and sorry if any of my calculations are wrong! you may want to double check all the calculations I've made in case of silly mistakes
I refer you to the Study Help Guidelines

https://www.thestudentroom.co.uk/hel...and-guidelines

Do not post full solutions unless absolutely necessary. Instead, try to guide the person asking towards working out the answer themself.
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username5333204
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#6
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(Original post by Pigster)
You are correct.

But I recommend not doing other people's work for them and instead try to guide people through a problem - people learn much better that way.
I should've just given prompts about the equations to use and to use the equation! Will make sure to do so in the future sorry about that
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Pigster
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#7
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(Original post by charco)
I refer you to the Study Help Guidelines

https://www.thestudentroom.co.uk/hel...and-guidelines

Do not post full solutions unless absolutely necessary. Instead, try to guide the person asking towards working out the answer themself.
wot he sed.
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