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20.0 cm3 of potassium hydroxide solution was titrated with 0.0500 M sulfuric acid. The titres were 12.4cm3, 12.3 cm3 and 12.2 cm3. What is the concentration of the potassium hydroxide in mol dm-3 and g dm-3?
Average titre = (12.40 + 12.30 + 12.20)/3 = 12.30 cm3
Number of moles of H2SO4 = 12.30/1000 × 0.0500
|
2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l) |
2KOH ºH2SO4
\ number of moles of KOH in 20.0 cm3 = 2 × 12.30/1000 × 0.0500
\concentration of KOH = (2 × 12.30/1000 × 0.0500)/0.020 = 0.062 mol dm-3
KOH = 56
\concentration of KOH = 0.062 × 56 = 3.4 g dm-3