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# Titration calculation watch

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1. In a reaction between a solution of a metallic hydroxide, MOH, and dilute hydrochloric acid, 20.0 cm3 if the alkali reacted with 25.0cm3 of the acid. The acid concentration was 4.00g per litre and that of the alkali 7.67g per litre.

Calculate:
a) the relative molecular mass of the alkali
b) the relative molecular mass of the metal, M
2. (Original post by 15076683)
In a reaction between a solution of a metallic hydroxide, MOH, and dilute hydrochloric acid, 20.0 cm3 if the alkali reacted with 25.0cm3 of the acid. The acid concentration was 4.00g per litre and that of the alkali 7.67g per litre.

Calculate:
a) the relative molecular mass of the alkali
b) the relative molecular mass of the metal, M
This means that the balanced equation for this reaction is:

1MOH + 1HCl --> MCl + H20

You can then deduce that one mole of the acid reacts with one mole of the unknown metallic alkali.

You have concentrations and volumes, by changing the units, you can find out the masses of the reacted acid and alkali.
Remember:
1 dm^3 = 1000 cm^3
1 litre = 1 dm^3
and n/v = c

Once you have the mass of your acid, you can find out how many moles of it reacted. This means, of course, you know how many moles of your unknown metallic hydroxide reacted.

With the two pieces of information, the moles and mass of the alkali, you should easily be able to find out the molecular mass.

M/mr = n

Hopefully this is helpful to you
3. (Original post by pineneedles)
This means that the balanced equation for this reaction is:

1MOH + 1HCl --> MCl + H20

You can then deduce that one mole of the acid reacts with one mole of the unknown metallic alkali.

You have concentrations and volumes, by changing the units, you can find out the masses of the reacted acid and alkali.
Remember:
1 dm^3 = 1000 cm^3
1 litre = 1 dm^3
and n/v = c

Once you have the mass of your acid, you can find out how many moles of it reacted. This means, of course, you know how many moles of your unknown metallic hydroxide reacted.

With the two pieces of information, the moles and mass of the alkali, you should easily be able to find out the molecular mass.

M/mr = n

Hopefully this is helpful to you
Thanks for your help! Much appreciated. However, I do have a question as I'm not sure how to turn the 'g per litre' values into moldm^-3.

If you know how to do this, could you please lend a hand. Thanks
4. (Original post by 15076683)
Thanks for your help! Much appreciated. However, I do have a question as I'm not sure how to turn the 'g per litre' values into moldm^-3.

If you know how to do this, could you please lend a hand. Thanks
Sure thing, so you have some volumes in cm^3, which, using the fact that 1 dm^3 = 1000cm^3, you can convert into dm^3.

25cm^3 = 0.025 dm^3

By multiplying the number of g per litre by this, you can find out the number of grams

so 0.025 x 4.00 = 0.1 g
Then you can find out the number of moles of acid by dividing this by the mr of HCl. You can turn this into moldm^3 but you don't really need to for this question.
5. (Original post by pineneedles)
Sure thing, so you have some volumes in cm^3, which, using the fact that 1 dm^3 = 1000cm^3, you can convert into dm^3.

25cm^3 = 0.025 dm^3

By multiplying the number of g per litre by this, you can find out the number of grams

so 0.025 x 4.00 = 0.1 g
Then you can find out the number of moles of acid by dividing this by the mr of HCl. You can turn this into moldm^3 but you don't really need to for this question.

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Updated: November 1, 2015
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