When matter is transformed from the liquid state to the gas state, the distance between the particles _______ and the attractive
1 answer:
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Explanation:
Molarity of the acetic acid = 0.250 M
Volume of the acetic acid solution = 10.0 mL = 0.010 L( 1 mL =0.001L)
Moles of acetic acid ;
Molarity of the NaOH = 0.200 M
a) Volume of the NaOH solution = 10.0 mL = 0.010 L( 1 mL =0.001L)
Moles of NaOH :
1 mole NaOH neutralizes 1 mole of acetic acid , then 0.002 moles of NaOH will neutralize 0.002 mol of acetic acid.
Moles of acetic acid left un-neutralized = 0.0025 mol - 0.002 = 0.0005 mol
1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0005 mole of acetic acid will give 0.0005 mole of hydrogen ions.
Moles of hydrogen ion= 0.0005 mol
Volume of the solution = 0.010 L+ 0.010 L = 0.020 L
The pH of the 10.0 mL of base added to acetic acid solution :
b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)
Moles of NaOH :
1 mole NaOH neutralizes 1 mole of acetic acid , then 0.0024 moles of NaOH will neutralize 0.0024 mol of acetic acid.
Moles of acetic acid left un-neutralized = 0.0025 mol - 0.0024 = 0.0001 mol
1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0001 mole of acetic acid will give 0.0001 mole of hydrogen ions.
Moles of hydrogen ion= 0.0001 mol
Volume of the solution = 0.010 L+ 0.012 L = 0.022 L
The pH of the 12.0 mL of base added to acetic acid solution :
c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)
Moles of NaOH :
1 mole NaOH neutralizes 1 mole of acetic acid , then 0.003 moles of NaOH will neutralize 0.003 mol of acetic acid.
All the moles of acetic acid will get neutralized by NaOH and un-neutralized sodium hydroxide will left over.
Moles of NaOH left un-neutralized = 0.003 mol - 0.0025 = 0.0005 mol
1 mole of NaOH gives 1 mole of hydroxide ion, then 0.0005 mole of NaOH acid will give 0.0005 mole of hydroxide ions.
Moles of hydroxide ion= 0.0005 mol
Volume of the solution = 0.010 L+ 0.015 L = 0.025 L
The pOH of the 15.0 mL of base added to acetic acid solution :
The pH of the 15.0 mL of base added to acetic acid solution :