Ezy concentration Q (Prelim) (1 Viewer)

Wohzazz

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I got this q wrong earlier in the year and still don't know where i went wrong- think i'm blind
5.0g of sodium hydroxide was dissolved in 200mL of water. Excess aluminium nitrate solution was added to the sodium hydroxide solution to precipitate all the hydroxide as aluminium hydroxide

edit/ - Calculate the concentration of the original hydroxide solution

I wrote:

(taking 200mL of water=200g)
Concentration= 5/200+5=2.4% (w/w)
 
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zeropoint

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NaOH(s) + H2O(l) ---> Na+(aq) + OH−(aq) + H2O(l)

Al(NO3)3(aq) + 3NaOH(aq) ---> Al(OH)3(s) + 3NaNO3(aq)

m(NaOH) = 5.0 g
n(NaOH) = m(NaOH)/M(NaOH)
= 5.0 g / 40 g mol^-1
= 0.13 mol
n(NaOH) = n(NaNO3)
m(NaNO3) = n(NaNO3)M(NaNO3)
= 0.125 mol * 85 g mol^-1
= 11 g

c(NaNO3) = m(NaNO3)/(m(NaNO3) + m(H2O))
= 10.625/(200 + 10.625)
= 5.0 %
 

spice girl

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express conc's in molar (moles/litre) unless stated otherwise

molar mass: MM(NaOH) = 22.99 + 16.00 + 1.008 = 40.00 g/mol
amount: n(NaOH) = 5.0/40.00 = 0.125 mol
volume of solution: V = 0.200 L
conc: [NaOH] = 0.125 / 0.200 = 0.625M (i.e. 0.625mol/L)
more correctly: [NaOH] = 0.63M (2 sig figs)
 

Wohzazz

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hmm..so what i did wrong was use the wrong concentration form ie. (mol/L) instead of (w/w), can they make you get zero for that?

can't believe i lost 2 marks for that
thank you spice girl.:)
 

zeropoint

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Originally posted by Wohzazz
hmm..so what i did wrong was use the wrong concentration form ie. (mol/L) instead of (w/w), can they make you get zero for that?
The term ``concentration'' refers to the amount-of-substance concentration; that is, the amount of substance divided by the volume. If the markers wanted the ``mass fraction'' they would have stated otherwise.
 

Wohzazz

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Well i'm not sure.. i was never sure with what concentrations to use anyway..guess i will work it out with the more question i used
What i think is that maybe the marker saw the answers sheet saying in (mol/L) got lazy and didn't bother to convert it to see if i got it right
 

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