I want to prepare the following solutions for titration;

0.50 Mol/dm3 HCl and
0.0250 mol/dm3 Na2CO3 (sodium trioxocarbonate(V)


Chemistry gurus please help.

What is amount of HCl that can be dissolved in a certain volume of water. Also, what is the amount of sodium trioxocarbonate (gram) that can be dissolved in certain amount of water. Thanks

@ Op go to play store and download molarity calculator. I believe it can solve your problem.

It's simple if I remember my chemistry correctly (volumetric analysis).

Using the formula

Number of moles= Concentration (C, in mol/dm3) * Volume (in cubic decimeters,dm3)
.

From your question, the concentration of HCL required is 0.5M.

Remember, number of moles (n) = mass/molar mass.
Therefore, we can rewrite the formula (above) as:

Mass/molar mass = C × V.

Molar mass of HCL = 1 + 35.5= 36.5g/mol

Since the concentration required is 0.5M, it means the volume is in dm3 which is also equivalent to a litre.

Consequently, our volume is 1dm3 or 1 litre.

what's left is the mass of HCL.

Thus, the mass of HCL required to give us 0.5M of HCL solution = molar mass of HCL× concentration × volume

= 36.5 × 0.5 × 1 = 18.25g.

In sum, we need to dissolve 18.25g of HCL in 1 litre of water (or any solution specified).

Please, follow suit and tackle that for sodium trioxocarbonate (IV).

Cheers.

YemziAdez:
It's simple if I remember my chemistry correctly (volumetric analysis).

Using the formula

Number of moles= Concentration (C, in mol/dm3) * Volume (in cubic decimeters,dm3)
.

From your question, the concentration of HCL required is 0.5M.

Remember, number of moles (n) = mass/molar mass.
Therefore, we can rewrite the formula (above) as:

Mass/molar mass = C × V.

Molar mass of HCL = 1 + 35.5= 36.5g/mol

Since the concentration required is 0.5M, it means the volume is in dm3 which is also equivalent to a litre.

Consequently, our volume is 1dm3 or 1 litre.

what's left is the mass of HCL.

Thus, the mass of HCL required to give us 0.5M of HCL solution = molar mass of HCL× concentration × volume

= 36.5 × 0.5 × 1 = 18.25g.

In sum, we need to dissolve 18.25g of HCL in 1 litre of water (or any solution specified).

Please, follow suit and tackle that for sodium trioxocarbonate (IV).

Cheers.

Thanks dear, I have also worked it out with the app someone mentioned above and it's giving me the same thing.


However, I still need some clarification specifically about the acid(HCl).

You know that the acid is in liquid state and should be measured using measuring cylinder?

Question: How do I measure it since the value you got above is in grams (36.5g)

Careartrust:


Thanks dear, I have also worked it out with the app someone mentioned above and it's giving me the same thing.


However, I still need some clarification specifically about the acid(HCl).

You know that the acid is in liquid state and should be measured using measuring cylinder?

Question: How do I measure it since the value you got above is in grams (36.5g)

Careartrust:


Thanks dear, I have also worked it out with the app someone mentioned above and it's giving me the same thing.


However, I still need some clarification specifically about the acid(HCl).

You know that the acid is in liquid state and should be measured using measuring cylinder?

Question: How do I measure it since the value you got above is in grams (36.5g)

Careartrust:


Thanks dear, I have also worked it out with the app someone mentioned above and it's giving me the same thing.


However, I still need some clarification specifically about the acid(HCl).

You know that the acid is in liquid state and should be measured using measuring cylinder?

Question: How do I measure it since the value you got above is in grams (36.5g)

convert the gram to ml ...hint:1000g=1l
1000ml=1L

Guess u want to do some sort of physiochemical analysis right?

Ysquare98:
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convert the gram to ml ...hint:1000g=1l
1000ml=1L

Guess u want to do some sort of physiochemical analysis right?

I just want to do Acid - Base titration for SSCE student