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HELP! Calculating the conc. of Sulphuric acid?

Long story short, i've been carrying out a chemistry investigation and at one point I had to vary the concentration of sulphuric acid.

Which i did by diluting my pur sulphuric acid using the following ratios:

1:29
1:19
1:9
1:4
1:3
1:1

Now that i'm writing up my investigation I need to convert these ratios into concentrations.

I would REALLY appreciate if someone could skim through my calculations and confirm whether what I'm doing is the correct way or a load of crap :smile:

NOTE: i'm using 4cm3 of the diluted acid in my experiment.

Formulas used:

•

Conc = moles/vol

•

moles = mass/RAM

•

mass = volume x density

Example of my calculation for 1:29 dilution (1 part sulphuric acid, 29 parts water)

1cm3+29cm3= 30cm3

(1/30) x 4cm3 = 2/15cm3 (The actual amount of Sulphuric acid in my 4cm3 of diluted solution)

Next calculating the moles, I need mass first!

so Mass= (2/15cm3) x 1.84g/cm3 = 0.245
(NOTE: 1.84G/cm3 is the density of pure sulphuric acid)

moles now,
moles = 0.245/98 = 0.0250mol

Conc:
0.0250mol/[(2/15cm3)/1000] = 18.7moldm-3

(edited 13 years ago)

Reply 1

Sifr

Yeah I'm definitely doing something wrong I keep getting 18.7molsdm-3 for all of them :/

Reply 2

joshed

What was your initial concentration of sulphuric acid?

Reply 3

Sifr

Original post by joshed

What was your initial concentration of sulphuric acid?

There is no initial conc!
Its 99% Pure sulphuric acid, I'm making up my own concentrations.

Reply 4

joshed

Well if it is pure of the pure it is around 18 molar sulphuric acid (I wouldn't spill it :P)

So your 1/30 dilution would be 18/30 molar solution so a 0.6M solution. (0.6 moldm^-3)

(edited 13 years ago)

Reply 5

Sifr

Original post by joshed

Well if it is pure of the pure it is around 18 molar sulphuric acid (I wouldn't spill it :P)

So your 1/30 dilution would be 18/30 molar solution so a 0.6M solution.