Enough water is added to the buffer in question 30 to make the total volume 5.00
ID: 229939 • Letter: E
Question
Enough water is added to the buffer in question 30 to make the total volume 5.00L. a) calculate the pH of the buffer B) calculate the pH of the buffer after adding .0250 mol of HCl to .376 of the buffer c) calculate the pH of the buffer after adding .0250 mol of KOH to .376 L of buffer d) comment on the effect of dilution on the pH of a buffer and on its buffer capacityQuestion 30 A buffer is made up of 239 mL of .187 M KHC4H4O6 and 137 mL of .288 M K2C4H4O6. Ka for H2C4H4O6 is 4.56x10^-5. Assuming volumes are additive, calculate. Enough water is added to the buffer in question 30 to make the total volume 5.00L. a) calculate the pH of the buffer B) calculate the pH of the buffer after adding .0250 mol of HCl to .376 of the buffer c) calculate the pH of the buffer after adding .0250 mol of KOH to .376 L of buffer d) comment on the effect of dilution on the pH of a buffer and on its buffer capacity
Question 30 A buffer is made up of 239 mL of .187 M KHC4H4O6 and 137 mL of .288 M K2C4H4O6. Ka for H2C4H4O6 is 4.56x10^-5. Assuming volumes are additive, calculate. a) calculate the pH of the buffer B) calculate the pH of the buffer after adding .0250 mol of HCl to .376 of the buffer c) calculate the pH of the buffer after adding .0250 mol of KOH to .376 L of buffer d) comment on the effect of dilution on the pH of a buffer and on its buffer capacity
Question 30 A buffer is made up of 239 mL of .187 M KHC4H4O6 and 137 mL of .288 M K2C4H4O6. Ka for H2C4H4O6 is 4.56x10^-5. Assuming volumes are additive, calculate.
Explanation / Answer
. initial pH of buffer in 30.
pH = pKa + log(base/acid)
= 4.34 + log[(39.456 mmol/44.693 mmoles)]
= 4.28
when it is diluted to volume = 5 L
initial volume of buffer = 239 + 137 = 376 ml
final [K2C4H4O6] = (0.288 M x 137 ml)/5000 ml = 0.0079 M
final [KHC4H4O6] = (0.187 M x 239 ml)/5000 ml = 0.0089 M
final pH
pH = 4.34 + log(0.0079/0.0089)
= 4.28
(b) after HCl = 0.025 mol added
new [K2C4H4O6] = 0.105 M x 0.376 L - 0.025 = 0.0145 mol
final [KHC4H4O6] = 0.120 M x 0.376 L + 0.025 = 0.0701 mol
final pH
pH = 4.34 + log(0.0145/0.0701)
= 3.65
(c) after KOH = 0.025 mol added
new [K2C4H4O6] = 0.105 M x 0.376 L + 0.025 = 0.0701 mol
final [KHC4H4O6] = 0.120 M x 0.376 L - 0.025 = 0.0145 mol
final pH
pH = 4.34 + log(0.0701/0.0145)
= 5.02
(d) dilution does not change the pH of the solution as ratio of conjugate base/acid remained the same.
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