greenamy2
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I have been struggling with this titration question and cannot find any solutions online, any help will be appreciated.



A student carries out a titration to find the concentration of 2AgOH. The equation for the reaction is: 2AgOH + H2SO4 --> Ag2SO4 + 2H2O. The students adds 25.0cm3 of AgOH solution into a conical flask and titrates it against 0.2 mol/dm3 of H2SO4 . The average titre is 8.13 cm3. Calculate the concentration of the AgOH
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EvelinaU
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The ratio of AgOH to H2SO4 is 2:1 and we need to use the formula moles (n) = concentration x volume

We can calculate the moles (n) of H2SO4 by multiplying the titre to the concentration. Remember to convert cm3 to dm3 so 8.13 cm3 turn to 0.00813 dm3.

0.00813 x 0.2 = 0.001626

Now, because of our 2:1 ratio we must multiply the moles of H2SO4 by 2 to get 0.001626 x 2 = 0.003252 moles of AgOH

Given that we know the volume of AgOH to be 25 cm3 > 0.025 dm3 we can use the formula to find the concentration.

0.003252 / 0.025 = 0.13008 (and then you round it to the sf required by the question).

Hope this helped
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greenamy2
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(Original post by EvelinaU)
The ratio of AgOH to H2SO4 is 2:1 and we need to use the formula moles (n) = concentration x volume

We can calculate the moles (n) of H2SO4 by multiplying the titre to the concentration. Remember to convert cm3 to dm3 so 8.13 cm3 turn to 0.00813 dm3.

0.00813 x 0.2 = 0.001626

Now, because of our 2:1 ratio we must multiply the moles of H2SO4 by 2 to get 0.001626 x 2 = 0.003252 moles of AgOH

Given that we know the volume of AgOH to be 25 cm3 > 0.025 dm3 we can use the formula to find the concentration.

0.003252 / 0.025 = 0.13008 (and then you round it to the sf required by the question).

Hope this helped
Thank you so much for your help!
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scimus63
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try here it may help a bit:

https://www.science-revision.co.uk/titrations.html
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