13/2017 If a buffer is pr epared using 0.75 mol CH,COOH and 0.45 molCH,Coo in 1

Question

13/2017 If a buffer is pr epared using 0.75 mol CH,COOH and 0.45 molCH,Coo in 1 liter of aqueous solution, is the buffering capacity of the solution great enough to maintain its pH if 10 gNaOH is added? a. Yes, the concentration of acetate ions in the solution is great enough to neutralize the hydroxide. b. Yes, the concentration of acetic acid in the solution is great enough to neutralize the hydroxide. c.No, the concentration of acetate ions in the solution is not high enough to neutralize the hydroxide. d. No, the concentration of acetic acid in the solution is not high enough to neutralize the hydroxide. e. No, a weak acid cannot neutralize a strong base. If a buffer is prepared using 0.5 mol CHyCOOH and 0.5 molCH,COO in 1 liter of aqueous solution, is the buffering capacity of the solution great enough to maintain its pH if 30 gNaOH is added? a. Yes, the concentration of acetate ions in the solution is great enough to neutralize the hydroxide. b. Yes, the concentration of acetic acid in the solution is great enough to neutralize the hydroxide. c. No, the concentration of acetate ions in the solution is not high enough to neutralize the hydroxide. d. No, the concentration of acetic acid in the solution is not high enough to neutralize the hydroxide. e. No, a weak acid cannot neutralize a strong base. If a buffer is prepared using 0.75 mol CH,COOH and 0.45 molCH,COo in 1 liter of aqueous solution, is the buffering capacity of the solution great enough to maintain its pH if 20 gHCI is added? a. Yes, the concentration of acetate ions in the solution is great enough to neutralize the acid. b. Yes, the concentration of acetic acid in the solution is great enough to neutralize the acid. c. No, the concentration of acetate ions in the solution is not high enough to neutralize the acid. d. No, the concentration of acetic acid in the solution is not high enough to neutralize the acid. e. No, a weak base cannot neutralize a strong acid. If a buffer is prepared using 0.55 mol CH3CHOHCOOH and 0.45 molCHjCHOHCOO in 1 liter of aqueous solution, is the buffering capacity of the solution great enough to maintain its pH if 5 gLiOH is added? a. Yes, the concentration of lactate ions in the solution is great enough to neutralize the hydroxide. b. Yes, the concentration of lactic acid in the solution is great enough to neutralize the hydroxide. e. No, the concentration of lactate ions in the solution is not high enough to neutralize the hydroxide. d. No, the concentration of lactic acid in the solution is not high enough to neutralize the hydroxide. . No, a weak acid cannot neutralize a strong base.

Explanation / Answer

1)

b) Yes, the concentration of acetic acid is enough to neutralize the hydroxide.

Ka= 1.8x10^-5

pKa = - log 1.8 x10^-5 = 4.74

Moles of CH3COOH = 0.75 moles

Moles of CH3COO- = 0.45 moles

Buffer is prepared in 1.0 L of aqueous solution, so there is no need to calculate concentration.

Concentration = Moles / volume in L

Concentration = Moles /1 = Concentration = Moles

By using Henderson–Hassel Balch equation, calculate pH of CH3COO– / CH3COOH buffer

pH = pKa + log [CH3COO– ] / [CH3COOH]

pH = 4.74 + log (0.45/0.75)

pH = 4.74 - 0.22 = 4.52

Calculate moles of NaOH added

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 10g/40g/mol = 0.25 mol

When 0.25mol NaOH is added to given buffer solution, then 0.25mol of CH3COOH reacts with 0.25mol NaOH and moles of CH3COOH left = 0.75 mol – 0.25 mol = 0.50 mol

When 0.25mol NaOH is added to given buffer solution, then 0.25mol of OH added to CH3COO–, so

CH3COO– = 0.45 mol + 0.25 mol = 0.70 mol

By using Henderson–Hassel Balch equation, calculate the pH of solution after addition of 0.25mol NaOH

pH = pKa + log [CH3COO– ] / [CH3COOH]

pH = 4.74 + log (0.70/0.50)

pH = 4.74 + 0.15 = 4.89

There is minor change in pH.

2) (d) No, the concentration of acetic acid is not high enough to neutralize the hydroxide.

Moles of CH3COOH = 0.50 moles

Moles of CH3COO- = 0.50 moles

Calculate moles of NaOH added

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 30g/40g/mol = 0.75 mol

When 0.25mol NaOH is added to given buffer solution, then 0.25mol of

There is 0.50 mol of CH3COOH which neutralizes only 0.50mol NaOH. Moles of NaOH left = 0.75 mol – 0.50 mol = 0.25 mol

So, the concentration of acetic acid is not high enough to neutralize the hydroxide

3)

c) No, the concentration of acetate is not high enough to neutralize the acid.

If a buffer is prepared using 0.75 mol HC2H3O2 and 0.45 mol C2H3O2- in 1.0 L of aqueous solution, is the buffering capacity of the solution great enough to maintain its pH if 20. g HCl is added

Moles of CH3COOH = 0.75 moles

Moles of CH3COO- = 0.45 moles

Calculate moles of HCl added

Molar mass of HCl = 36.46 g/mol

Moles of HCl = 20gHCl/36.46 g/mol = 0.55 mol

Mol of HCl is greater than moles of CH3COO-, so it is not enough to neutralize the acid.

4) (b) Yes, the concentration of lactic acid is great enough to neutralize the hydroxide.

Calculate moles of LiOH added

Molar mass of LiOH = 23.95 g/mol

Moles of LiOH = 5.0g/23.95g/mol = 0.21 mol

Mol CH3CHOHCOOH = 0.55

1)

b) Yes, the concentration of acetic acid is enough to neutralize the hydroxide.

Ka= 1.8x10^-5

pKa = - log 1.8 x10^-5 = 4.74

Moles of CH3COOH = 0.75 moles

Moles of CH3COO- = 0.45 moles

Buffer is prepared in 1.0 L of aqueous solution, so there is no need to calculate concentration.

Concentration = Moles / volume in L

Concentration = Moles /1 = Concentration = Moles

By using Henderson–Hassel Balch equation, calculate pH of CH3COO– / CH3COOH buffer

pH = pKa + log [CH3COO– ] / [CH3COOH]

pH = 4.74 + log (0.45/0.75)

pH = 4.74 - 0.22 = 4.52

Calculate moles of NaOH added

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 10g/40g/mol = 0.25 mol

When 0.25mol NaOH is added to given buffer solution, then 0.25mol of CH3COOH reacts with 0.25mol NaOH and moles of CH3COOH left = 0.75 mol – 0.25 mol = 0.50 mol

When 0.25mol NaOH is added to given buffer solution, then 0.25mol of OH added to CH3COO–, so

CH3COO– = 0.45 mol + 0.25 mol = 0.70 mol

By using Henderson–Hassel Balch equation, calculate the pH of solution after addition of 0.25mol NaOH

pH = pKa + log [CH3COO– ] / [CH3COOH]

pH = 4.74 + log (0.70/0.50)

pH = 4.74 + 0.15 = 4.89

There is minor change in pH.

2) (d) No, the concentration of acetic acid is not high enough to neutralize the hydroxide.

Moles of CH3COOH = 0.50 moles

Moles of CH3COO- = 0.50 moles

Calculate moles of NaOH added

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 30g/40g/mol = 0.75 mol

When 0.25mol NaOH is added to given buffer solution, then 0.25mol of

There is 0.50 mol of CH3COOH which neutralizes only 0.50mol NaOH. Moles of NaOH left = 0.75 mol – 0.50 mol = 0.25 mol

So, the concentration of acetic acid is not high enough to neutralize the hydroxide

3)

c) No, the concentration of acetate is not high enough to neutralize the acid.

If a buffer is prepared using 0.75 mol HC2H3O2 and 0.45 mol C2H3O2- in 1.0 L of aqueous solution, is the buffering capacity of the solution great enough to maintain its pH if 20. g HCl is added

Moles of CH3COOH = 0.75 moles

Moles of CH3COO- = 0.45 moles

Calculate moles of HCl added

Molar mass of HCl = 36.46 g/mol

Moles of HCl = 20gHCl/36.46 g/mol = 0.55 mol

Mol of HCl is greater than moles of CH3COO-, so it is not enough to neutralize the acid.

4) (b) Yes, the concentration of lactic acid is great enough to neutralize the hydroxide.

Calculate moles of LiOH added

Molar mass of LiOH = 23.95 g/mol

Moles of LiOH = 5.0g/23.95g/mol = 0.21 mol

Mol CH3CHOHCOOH = 0.55

Mol CH3CHOHCOOH is enough to neutralize the hydroxide.

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