Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid
ID: 891371 • Letter: P
Question
Part A
What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.507 mol of NaA in 2.00 L of solution? The dissociation constant Kaof HA is 5.66×107.
Express the pH numerically to three decimal places.
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Part B
What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.
Express the pH numerically to three decimal places.
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Part C
What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.
Express the pH numerically to three decimal places.
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pH =Explanation / Answer
moles of weak acid = 0.809
moles of NaA = 0.507
Ka of HA = 5.66×10^7
PART A:
pKa = 6.25
HA + NaOH ------------------> NaA + H2O
pH = pKa + log[salt / acid]
= 6.25 + log[0.507 /0.809]
= 6.05
pH = 6.05
PART B :
0.15 moles of HCl added to above buffer.
pH = pKa + log [Salt –C/acid + C]
= 6.25 + log [0.507 - 0.15 / 0.809 + 0.15]
= 5.82
pH = 5.82
PART C:
on addition of 0.195 mol of NaOH to buffer
on addition of ’ C’ moles of base to acidic buffer salt moles increases and acid moles decreases
so
pH = pKa + log [Salt +C/acid - C]
= 6.25 + log [0.507 + 0.195 / 0.809 - 0.195]
= 6.31
pH = 6.31