### More Titration Q - SG Chem

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• #### Message 1.

Posted by MissChemTeacher_Bitesize (U15172316) on Thursday, 10th January 2013

Hey!

Here are some titration questions you were after. If you need help answering any of them get in touch.

1) A titration found that 20.2cm³ of sodium hydroxide was neutralised by 15.5cm³ of sulphuric acid with a concentration of 0.1 mol/l.

Use this information and the balanced equation to calculate the concentration of sodium hydroxide.

H₂SO₄ + 2NaOH ----------> Na₂SO₄ + 2H₂O

2) In a titration 10cm³ of potassium hydroxide solution was neutralised by 12cm³ of 1.2mol/l nitric acid.
Use this information and balanced equation to calculate the concentration of potassium hydroxide.

KOH + HNO₃ ----------> KNO₃ + H₂O

Good Luck with these questions Miss_Chem

• #### Message 2

, in reply to message 1.

Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 21st January 2013

Hey Guys!
Here are the solutions to these titration problems.

H₂SO₄ + 2NaOH ----------> Na₂SO₄ + 2H₂O

n of moles of H₂SO₄ = concentration x volume in litres
n of moles of H₂SO₄ = 0.1 x 15.5/1000
n of moles of H₂SO₄ = 0.00155 moles

Use the ratio: H₂SO₄ ……:…….2NaOH
Mole Ratio:………1………….:………..2
n from above:..0.0015..:…..0.0031 moles

concentration of NaOH = n of moles / volume in litres
concentration of NaOH = 0.0031 / 0.01 litres
concentration of NaOH = 0.153 mol/l

• #### Message 3

, in reply to message 2.

Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 21st January 2013

Whoops! The volume of NaOH should've been 0.0202l

• #### Message 4

, in reply to message 3.

Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 21st January 2013

Here is the solution to question 2

KOH + HNO₃ ----------> KNO₃ + H₂O

n of moles of HNO₃ = concentration x volume in litres
n of moles of HNO₃ = 1.2 x 12/1000
n of moles of HNO₃ = 0.0144 moles

Use the ratio: HNO₃ : KOH
Mole Ratio:…….1……:……1
n from above: 0.0144: 0.0144

concentration of KOH = n of moles / volume in litres
concentration of KOH = 0.0144 / 0.01
concentration of KOH = 1.44 mol/l

Hope these solutions helped you out.

Cheers Miss_Chem

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