Answer :
When 59.1 ml of 0.151 M [tex]NaOH[/tex] was needed to neutralize 50.0 ml of an [tex]H_2SO_4[/tex] solution, the concentration of the original sulfuric acid solution is 0.17856 M.
The balanced chemical equation for the reaction between [tex]NaOH[/tex] and [tex]H_2SO_4[/tex] is:
[tex]NaOH (aq) + H_2SO_4 (aq)[/tex] → [tex]Na_2SO_4 (aq) + 2H_2O (l)[/tex]
From the equation, we can see that one mole of [tex]NaOH[/tex] reacts with one mole of [tex]H_2SO_4[/tex]. Therefore, the number of moles of [tex]H_2SO_4[/tex] in the original solution is equal to the number of moles of [tex]NaOH[/tex] used to neutralize it.
The volume of NaOH used is 59.1 ml, and the molarity of [tex]NaOH[/tex] is 0.151 M. Using the formula for calculating the number of moles:
moles of [tex]NaOH[/tex] = molarity x volume (in liters)
moles of [tex]NaOH[/tex] [tex]=0.151 * 0.0591[/tex]
moles of [tex]NaOH[/tex] = 0.008929 moles
Since one mole of [tex]H_2SO_4[/tex] reacts with one mole of [tex]NaOH[/tex], the number of moles of [tex]H_2SO_4[/tex] in the original solution is also 0.008929 moles.
The volume of the original [tex]H_2SO_4[/tex] solution is 50.0 ml, or 0.0500 L. Using the formula for molarity:
Molarity of [tex]H_2SO_4[/tex] = moles of [tex]H_2SO_4[/tex] / volume (in liters)
Molarity of [tex]H_2SO_4[/tex] [tex]= \frac{0.008929}{0.0500}[/tex]
Molarity of [tex]H_2SO_4[/tex] = 0.17856 M
Hence, the primordial sulfuric acid solution has a concentration of 0.17856 M.
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