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1.973 g of impure potassium nitrate was reacted with 2 g of aluminium (an excess) and 25 cm3 of 2 M sodium hydroxide solution (an excess) according to the following equation:
|
3NO3-(aq) + 8Al(s) + 5OH-(aq) + 2H2O(l) ⇒ 8AlO2-(aq) + 3NH3(g) |
The reaction was carried out in a sealed container with the evolved ammonia gas being dissolved in 25.0 cm3 of 1.000 M hydrochloric acid through an inverted funnel. The original reaction mixture was briefly boiled. The excess hydrochloric acid was then titrated with 0.500 M sodium hydroxide solution. The titre was 12.60 cm3.
What is the 25 cm3 of 2 M sodium hydroxide solution measured out with?
Why was the original reaction mixture briefly boiled?
What is the percentage purity of the potassium hydroxide?
A measuring cylinder can be used to measure out the 2 M sodium hydroxide solution. It is an excess, so the volume is not critical and not used in the calculation.
The original reaction mixture is briefly boiled so that any dissolved ammonia is driven out of the solution.
Number of moles of NaOH = 0.500 × 12.60/1000
|
HCl(aq) + NaOH(aq) ⇒ NaCl(aq) + H2O(l) |
HCl º NaOH
\ Number of moles of HCl titrated = 0.500 × 12.60/1000 = 0.00630
Number of moles of HCl originally = 1.00 × 25.0/1000 = 0.0250
\ Number of moles of HCl reacted with NH3 = 0.0250 - 0.00630 = 0.0187
|
HCl(aq) + NH3(aq) ⇒ NH4Cl(aq) |
NH3 º HCl
\ Number of moles of NH3 from original reaction = 0.0187
|
3NO3-(aq) + 8Al(s) + 5OH-(aq) + 2H2O(l) ⇒ 8AlO2-(aq) + 3NH3(g) |
NO3- º NH3
\ Number of moles of NO3- = 0.0187
KNO3 = 101
\ Mass of KNO3 = 0.0187 × 101 = 1.889 g
\ Percentage purity = 1.889/1.973 × 100 = 95.7%