Q90E
Question
What volume of a 0.00945 M solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a H2SO4 concentration of 1.23×10-4M
\({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2KO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ }}{{\rm{K}}_2}{\rm{S}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2}}{{\rm{H}}_2}{{\rm{O}}_{{\rm{(l)}}}}\)
Step-by-Step Solution
Verified1.30 ml of potassium hydroxide will be required to titrate 50.00 mL of acid rain sample with sulphuric acid.
It is given that the concentration of sulphuric acid is 1.23×10-4M and the volume is 50 ml.
The number of moles can be calculated by the mole concept formula that is:
the number of moles = \({\rm{volume }} \times {\rm{ concentration}}\)
the number of moles = 50×1.23×10-4M =6.15×10-3 moles
Now, as 1 moles of sulphuric acid reacts with 2 moles of \({\rm{KOH}}\), thus the number of moles of potassium hydroxide will be 6.15×10-3 moles×2=0.0123 moles
It is given that the concentration of potassium hydroxide is 0.00945 M and the moles of \({\rm{KOH}}\)from above calculation is 0.0123 moles, then the volume of \({\rm{KOH}}\) will be calculated as
\({\rm{Volume = }}\frac{{{\rm{moles}}}}{{{\rm{concentration}}}}\)
\({\rm{Volume = }}\frac{{0.0123}}{{0.00945}}{\rm{ = 1}}{\rm{.30 mL}}\)