Explanation:
![Molarity=\frac{moles}{Volume(L)}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7Bmoles%7D%7BVolume%28L%29%7D)
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 ;
![n=0.250 M\times 0.010 L=0.0025 mol](https://tex.z-dn.net/?f=n%3D0.250%20M%5Ctimes%200.010%20L%3D0.0025%20mol)
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 : ![0.200M\times 0.010 L=0.002 mol](https://tex.z-dn.net/?f=0.200M%5Ctimes%200.010%20L%3D0.002%20mol)
![CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O](https://tex.z-dn.net/?f=CH_3COOH%2BNaOH%5Crightarrow%20CH_3COONa%2BH_2O)
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
![[H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B0.0005%20mol%7D%7B0.020%20L%7D%3D0.025%20M)
The pH of the 10.0 mL of base added to acetic acid solution :
![pH=-\log[H^+]=-\log[0.025 M]=1.60](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.025%20M%5D%3D1.60)
b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)
Moles of NaOH : ![0.200M\times 0.012 L=0.0024 mol](https://tex.z-dn.net/?f=0.200M%5Ctimes%200.012%20L%3D0.0024%20mol)
![CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O](https://tex.z-dn.net/?f=CH_3COOH%2BNaOH%5Crightarrow%20CH_3COONa%2BH_2O)
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
![[H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B0.0001%20mol%7D%7B0.022%20L%7D%3D0.0045%20M)
The pH of the 12.0 mL of base added to acetic acid solution :
![pH=-\log[H^+]=-\log[0.0045 M]=2.34](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.0045%20M%5D%3D2.34)
c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)
Moles of NaOH : ![0.200M\times 0.015 L=0.003 mol](https://tex.z-dn.net/?f=0.200M%5Ctimes%200.015%20L%3D0.003%20mol)
![CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O](https://tex.z-dn.net/?f=CH_3COOH%2BNaOH%5Crightarrow%20CH_3COONa%2BH_2O)
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
![[OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B0.0005%20mol%7D%7B0.025%20L%7D%3D0.02%20M)
The pOH of the 15.0 mL of base added to acetic acid solution :
![pOH=-\log[OH^-]=-\log[0.02 M]=1.70](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B0.02%20M%5D%3D1.70)
The pH of the 15.0 mL of base added to acetic acid solution :
![pH=14-pOH=14-1.70=12.3](https://tex.z-dn.net/?f=pH%3D14-pOH%3D14-1.70%3D12.3)