Answer to Question #141671 in General Chemistry for sarah

Question

Part A) a buffer solution is made by adding 30.8 g of sodium acetate (NaCH3COO, molar mass= 82.03g/mol) to 250mL of 1.0M acetic acid(CH3COOH, Ka+ 1.8x10^-5). Assume no volume change. what is the pH of the buffer solution?
Part B) A reaction is conducted in the buffer solution in Part A that produces 0.075 mol of H3O+. what is the pH of the buffer solution after the reaction is complete?

Answer 1

Q141671

Part A) a buffer solution is made by adding 30.8 g of sodium acetate (NaCH3COO, molar mass= 82.03g/mol) to 250mL of 1.0M acetic acid(CH3COOH, Ka+ 1.8x10^-5). Assume no volume change. what is the pH of the buffer solution?

Solution :

Part A

Step 1 : To find the concentration of CH3COO^- and CH3COOH after mixing

Acetate ion and acetic acid will form a buffer system.

We will convert 30.8g of sodium acetate to moles using the given molar mass 82.03g/mol

moles of sodium acetate = 30.8g of NaCH₃COO * 1 mol NaCH₃COO/ 82.03 g NaCH₃COO

= 0.3755 mol NaCH₃COO

NaCH₃COO dissociates completely in water to give Na⁺ and CH₃COO^- .

1 NaCH₃COO(aq) ------> 1 Na⁺ (aq) + 1 CH₃COO^- (aq)

moles of acetate in the solution = 0.3755 mol NaCH₃COO * 1 mol CH₃COO^- /1 mol NaCH₃COO

= 0.3755 mol of CH₃COO^-

Molarity = moles of solute/Volume of solution in ‘L’

For finding the concentration of CH₃COO^- we will have to convert 250mL to ‘L’ using

1L = 1000mL

Volume of solution in ‘L’ = 250 mL * 1L/1000 mL = 0.250 L

[CH₃COO^- ] = 0.3755 mol of CH₃COO^- / 0.250 L = 1.502 mol/L = 1.502M

concentration of acetic acid ( CH₃COOH) is already given = 1.0 M

Step 2 : To find the pKa from Ka = 1.8x10^-5

pKa = -log[Ka] = log [ 1.8x10⁵ ] = 4.744;

Step 3 : To find the pH of the buffer using Henderson–Hasselbalch equation.

pH = pKa + log { [A^- ] / [HA] }

where [A-] is the concentration of basic species and [HA] is the concentration of acidic species in the buffer solution.

In the given buffer CH₃COO^- is the base and CH₃COOH is the acid

We know pKa = 4.744, [CH₃COO^- ] = 1.502M and [ CH₃COOH] = 1.0M.

Substitute this in the Henderson–Hasselbalch equation.

pH = 4.744 + log { 1.502M/ 1.0M }

pH = 4.744 + log (1.502) = 4.744 + 0.17667 = 4.9207.

Hence the pH of the buffer solution would be 4.92

Part B) A reaction is conducted in the buffer solution in Part A that produces 0.075 mol of H3O+. what is the pH of the buffer solution after the reaction is complete?

Part B :

The reaction is conducted in the buffer solution which produces 0.075 mol of H₃O⁺ . This

H₃O⁺ will combine with the base component ( acetate ion ) of the buffer and give the acetic acid.

CH₃COO^- (aq) + H₃O⁺ (aq) -----> CH₃COOH(aq) + H₂O (l)

concentration of H₃O⁺ produced from the reaction = 0.075 mol of H₃O⁺ / 0.250 L

= 0.300mol/L

The concentration of the H₃O⁺ from the buffer is small compared to that formed from the reaction.

So we can neglect the H₃O⁺ from the buffer.

We will have to draw the ICF table for this reaction.

CH₃COO^- (aq) + H₃O⁺ (aq) -----> CH₃COOH(aq) + H₂O (l)

Initial 1.502M 0.300M 1.0M.

Change - 0.300M - 0.300M +0.300M

Final 1.202M 0 1.300M

now substitute, [CH₃COO^- ] = 1.202M and [CH₃COOH] = 1.300M in the

Henderson–Hasselbalch equation we have

pH = 4.744 + log{ [CH₃COO^- ] / [CH₃COOH] }

pH = 4.744 + log{ 1.202M / 1.300M } = 4.744 + log (0.9246) = 4.744 – 0.0340

pH = 4.70996

pH = 4.71;

So the pH of the buffer solution after the completion of reaction is 4.71;