moles of base, using molar mass of NaOH 15.0g NaOH @ 40.0gmol = .pdf

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annammalassociates

moles of base, using molar mass of NaOH: 15.0g NaOH @ 40.0g/mol = 0.375 moles of NaOH moles of acid: 0.1500 litres @ 0.250moles/litre= 0.0375 moles of acid the reaction is 1:1 1 NaOH + 1 HNO3 --> NaNO3 + H2O so it uses up all of 0.0375 moles of acid and only destroys 10% of the Base, so it is still BASIC: 0.375 moles of NaOH - 0.0375 moles destroyed by the moles of acid .... & leaves behind 90% of the base , which is: 0.3375 moles base ------------ concentrations of all the ions present 0.3375 moles base left over, releases 0.3375 moles of OH- /0.150 litres... OH- = 2.25 molar Na + was not destroyed, it was a spectator ion ... Na+ stays the same @ 0.375 moles of NaOH / 0.150litres = 2.50 molar NO3- was not destroyed, it was a spectator ion ... NO3- stays the same @ 0.250 molar Solution moles of base, using molar mass of NaOH: 15.0g NaOH @ 40.0g/mol = 0.375 moles of NaOH moles of acid: 0.1500 litres @ 0.250moles/litre= 0.0375 moles of acid the reaction is 1:1 1 NaOH + 1 HNO3 --> NaNO3 + H2O so it uses up all of 0.0375 moles of acid and only destroys 10% of the Base, so it is still BASIC: 0.375 moles of NaOH - 0.0375 moles destroyed by the moles of acid .... & leaves behind 90% of the base , which is: 0.3375 moles base ------------ concentrations of all the ions present 0.3375 moles base left over, releases 0.3375 moles of OH- /0.150 litres... OH- = 2.25 molar Na + was not destroyed, it was a spectator ion ... Na+ stays the same @ 0.375 moles of NaOH / 0.150litres = 2.50 molar NO3- was not destroyed, it was a spectator ion ... NO3- stays the same @ 0.250 molar.

moles of base, using molar mass of NaOH:
15.0g NaOH @ 40.0g/mol = 0.375 moles of NaOH
moles of acid:
0.1500 litres @ 0.250moles/litre= 0.0375 moles of acid
the reaction is 1:1
1 NaOH + 1 HNO3 --> NaNO3 + H2O
so it uses up all of 0.0375 moles of acid and only destroys 10% of the Base, so it is still BASIC:
0.375 moles of NaOH - 0.0375 moles destroyed by the moles of acid .... & leaves behind 90% of
the base , which is: 0.3375 moles base
------------
concentrations of all the ions present
0.3375 moles base left over, releases 0.3375 moles of OH- /0.150 litres... OH- = 2.25 molar
Na + was not destroyed, it was a spectator ion ... Na+ stays the same @ 0.375 moles of NaOH /
0.150litres = 2.50 molar
NO3- was not destroyed, it was a spectator ion ... NO3- stays the same @ 0.250 molar
Solution
moles of base, using molar mass of NaOH:
15.0g NaOH @ 40.0g/mol = 0.375 moles of NaOH
moles of acid:
0.1500 litres @ 0.250moles/litre= 0.0375 moles of acid
the reaction is 1:1
1 NaOH + 1 HNO3 --> NaNO3 + H2O
so it uses up all of 0.0375 moles of acid and only destroys 10% of the Base, so it is still BASIC:

0.375 moles of NaOH - 0.0375 moles destroyed by the moles of acid .... & leaves behind 90% of
the base , which is: 0.3375 moles base
------------
concentrations of all the ions present
0.3375 moles base left over, releases 0.3375 moles of OH- /0.150 litres... OH- = 2.25 molar
Na + was not destroyed, it was a spectator ion ... Na+ stays the same @ 0.375 moles of NaOH /
0.150litres = 2.50 molar
NO3- was not destroyed, it was a spectator ion ... NO3- stays the same @ 0.250 molar

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moles of base, using molar mass of NaOH 15.0g NaOH @ 40.0gmol = .pdf

1. moles of base, using molar mass of NaOH: 15.0g NaOH @ 40.0g/mol = 0.375 moles of NaOH moles of acid: 0.1500 litres @ 0.250moles/litre= 0.0375 moles of acid the reaction is 1:1 1 NaOH + 1 HNO3 --> NaNO3 + H2O so it uses up all of 0.0375 moles of acid and only destroys 10% of the Base, so it is still BASIC: 0.375 moles of NaOH - 0.0375 moles destroyed by the moles of acid .... & leaves behind 90% of the base , which is: 0.3375 moles base ------------ concentrations of all the ions present 0.3375 moles base left over, releases 0.3375 moles of OH- /0.150 litres... OH- = 2.25 molar Na + was not destroyed, it was a spectator ion ... Na+ stays the same @ 0.375 moles of NaOH / 0.150litres = 2.50 molar NO3- was not destroyed, it was a spectator ion ... NO3- stays the same @ 0.250 molar Solution moles of base, using molar mass of NaOH: 15.0g NaOH @ 40.0g/mol = 0.375 moles of NaOH moles of acid: 0.1500 litres @ 0.250moles/litre= 0.0375 moles of acid the reaction is 1:1 1 NaOH + 1 HNO3 --> NaNO3 + H2O so it uses up all of 0.0375 moles of acid and only destroys 10% of the Base, so it is still BASIC:

2. 0.375 moles of NaOH - 0.0375 moles destroyed by the moles of acid .... & leaves behind 90% of the base , which is: 0.3375 moles base ------------ concentrations of all the ions present 0.3375 moles base left over, releases 0.3375 moles of OH- /0.150 litres... OH- = 2.25 molar Na + was not destroyed, it was a spectator ion ... Na+ stays the same @ 0.375 moles of NaOH / 0.150litres = 2.50 molar NO3- was not destroyed, it was a spectator ion ... NO3- stays the same @ 0.250 molar