Answer :
Explanation:
[tex]Molarity=\frac{moles}{Volume(L)}[/tex]
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 ;
[tex]n=0.250 M\times 0.010 L=0.0025 mol[/tex]
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 : [tex]0.200M\times 0.010 L=0.002 mol[/tex]
[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex]
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
[tex][H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M[/tex]
The pH of the 10.0 mL of base added to acetic acid solution :
[tex]pH=-\log[H^+]=-\log[0.025 M]=1.60[/tex]
b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)
Moles of NaOH : [tex]0.200M\times 0.012 L=0.0024 mol[/tex]
[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex]
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
[tex][H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M[/tex]
The pH of the 12.0 mL of base added to acetic acid solution :
[tex]pH=-\log[H^+]=-\log[0.0045 M]=2.34[/tex]
c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)
Moles of NaOH : [tex]0.200M\times 0.015 L=0.003 mol[/tex]
[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex]
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
[tex][OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M[/tex]
The pOH of the 15.0 mL of base added to acetic acid solution :
[tex]pOH=-\log[OH^-]=-\log[0.02 M]=1.70[/tex]
The pH of the 15.0 mL of base added to acetic acid solution :
[tex]pH=14-pOH=14-1.70=12.3[/tex]