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;
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