Chemistry, 25.02.2020 19:47 kaylailkanic1487

Be sure to answer all parts. For the titration of 10.0 mL of 0.250 M acetic acid with 0.200 M sodium hydroxide, determine the pH when: (a) 10.0 mL of base has been added. (b) 12.5 mL of base has been added. (c) 15.0 mL of base has been added.

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Answer from: jhonnysoriano9053

Explanation:

$Molarity=\frac{moles}{Volume(L)}$

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 ;

$n=0.250 M\times 0.010 L=0.0025 mol$

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 : $0.200M\times 0.010 L=0.002 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

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

$[H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M$

The pH of the 10.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.025 M]=1.60$

b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.012 L=0.0024 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

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

$[H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M$

The pH of the 12.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.0045 M]=2.34$

c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.015 L=0.003 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

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

$[OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M$

The pOH of the 15.0 mL of base added to acetic acid solution :

$pOH=-\log[OH^-]=-\log[0.02 M]=1.70$

The pH of the 15.0 mL of base added to acetic acid solution :

pH=14-pOH=14-1.70=12.3

Answer from: Quest

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.

Answer from: Quest

the answer is d. 74.02

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Be sure to answer all parts. For the titration of 10.0 mL of 0.250 M acetic acid with 0.200 M sodium...