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Full answer.

$\text{Let }E_B\text{ be the hydroxide ions of the basic effluent}\\ \text{and }E_A\text{ be the hydrogen ions of the acidic effluent.}$
It's not really good notation. I just realised I stuffed up halfway through typing and was too lazy to fix it.

We are only interested in the formula $C = \frac{n}{V} $
$\text{For }E_B\text{ we note that }n_{H_2SO_4} = 0.0295 \times 0.233 = 6.8375\times 10^{-3}\text{ mol}.\\ \text{Thus, the moles of the acid is }n_{H^+} = 1.3747\times 10^{-2}\text{ mol}\\ \text{so by consequence we also have }n_{E_B} = 1.3747\times 10^{-2}\text{ mol.}$
$\text{This implies that }E_B\text{ has concentration }[E_B] = \frac{1.3747\times 10^{-2}}{0.025} = 0.54988\text{ mol L}^{-1}.$
Again, a remark that they have exactly neutralised each other. Else we would not be able to make such a deduction.
$\text{For }E_A\text{ we note that }n_{NaOH} = 0.0328 \times 0.274 = 8.9872\times 10^{-3}\text{ mol}.\\ \text{But of course this means the moles of the base is }n_{OH^-} = 8.9872\times 10^{-3}\text{ mol}\\ \text{and hence }n_{E_A} = 8.9872\times 10^{-3}\text{ mol.}$
$\text{This implies that }E_A\text{ has concentration }[E_A] = \frac{8.9872\times 10^{-3}}{0.025}=0.359488\text{ mol}.$

$\text{Now, reacting between }E_A\text{ and }E_B\text{ is just ordinary neutralisation (again).}\\ \text{So if we want the neutralisation to be exact, we need equal moles of both.}$
$\text{Given that we have 100L of }E_A,\\ n_{E_A} = 359.488 \text{ mol}.$
$\text{For the exactedness, we need this amount of the basic effluent as well.}\\ \text{So }n_{E_B} = 359.488\text{, and hence }V_{E_B} = \frac{35.9488}{0.54988} = 65.375718\dots \text{ L}.$
Which rounds to the given answer.

ATAR Notes: Forum

HSC Stuff => HSC Science Stuff => HSC Subjects + Help => HSC Chemistry => Topic started by: Lfex on April 10, 2018, 04:19:59 pm