What volume (in mL) of .158 M KOH solution is needed to titrate 20.0 mL of .0903 M H2SO4 to the equivalance point?

Jul 3, 2017

Approx. a $23 \cdot m L$ volume of the sulfuric acid is required...........

Explanation:

We need (i), a stoichiometric equation.......

${H}_{2} S {O}_{4} \left(a q\right) + 2 K O H \left(a q\right) \rightarrow {K}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

And (ii) equivalent quantities of $\text{sulfuric acid}$,

$\text{Moles of sulfuric acid} = 20.0 \cdot \cancel{m L} \times {10}^{-} 3 \cdot \cancel{L} \cdot \cancel{m {L}^{-} 1} \times 0.0903 \cdot m o l \cdot \cancel{{L}^{-} 1} = 1.806 \times {10}^{-} 3 \cdot m o l .$

And, CLEARLY, we need $\text{2 equiv potassium hydroxide......}$

$\frac{2 \times 1.806 \times {10}^{-} 3 \cdot m o l}{0.158 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 22.9 \cdot m L .$

This is a good question because in the laboratory we would typically LIKE to deal with a titration whose TOTAL volume was around about $50 \cdot m L$.